= Solution
Use an oriented <geodesic> hyperbolic $2s$-gon $D$ with paired edges of equal length. Identify each pair by a length-preserving map reversing the <boundary> directions, to make an oriented base surface $M_0$ with possibly singular vertices. Assign to the $s$ edge pairs generators $A_1,\ldots,A_s$ of $T$, with inverse labels for reversed crossings. For each $g\in T$, take a copy $D_g$ and glue its positively labelled edge of type $k$ to the corresponding mate in $D_{gA_k}$. The generators make the glued surface $M$ connected. Left multiplication $D_g\mapsto D_{hg}$ commutes with these right-labelled gluings and gives the $T$ action with quotient $M_0$.
Existence for an arbitrary finite <group> can even be arranged with no cone singularities. If $T$ has $d$ generators, take $h\ge\max\{2,d\}$ and a regular <hyperbolic polygon> with $4h$ edges, all angles $\pi/(2h)$, paired with side word $\prod_{i=1}^h a_ib_ia_i^{-1}b_i^{-1}$. Its vertices form one cycle of angle $2\pi$, giving a smooth base of <genus> $h$. Label the $a_i$ by the generators of $T$, padded by identities, and every $b_i$ by the identity. The single vertex word is $\prod_i[A_i,1]=1$, so every lifted vertex also has angle $2\pi$. The resulting connected ordinary <normal covering map> has degree $|T|$ and deck <group> $T$.
This is the <paired-polygon construction of finite surface covers>. Interior and open-edge points are regular hyperbolic points; possible singularities occur only at vertices. When the base has cone points, the map is an <orbifold> covering and may be branched as a map between the underlying topological surfaces. It should not be called an unbranched <manifold> covering at those branch points.
A <vertex cycle of a paired polygon> is the collection of vertices identified to one point on $M_0$. Let $r$ be the number of cycles, $\theta_j$ the sum of the polygon angles in cycle $j$, and $C_j\in T$ the gluing word recorded on going around it. If $q_j=|C_j|$ is the order of that element, a lift closes after $q_j$ turns. It has angle $q_j\theta_j$, and there are $|T|/q_j$ such lifted vertices. The base and lifted cone orders, when they are integer <orbifold> orders, are
$$
\boxed{m_j=\frac{2\pi}{\theta_j},\qquad \widetilde m_j=\frac{2\pi}{q_j\theta_j}=\frac{m_j}{q_j}.}
$$
An <orbifold> homomorphism requires $q_j\mid m_j$. The lifted point is smooth exactly when $q_j\theta_j=2\pi$, equivalently $q_j=m_j$ in that case. For arbitrary cone metrics the angle formulas remain valid even when an angle does not correspond to an integer <orbifold> order.
Counting faces, edges and vertices gives the <Euler characteristic of a paired-polygon surface cover>:
$$
\boxed{\chi(M_0)=1-s+r,\qquad \chi(M)=|T|\left(1-s+\sum_{j=1}^r\frac1{q_j}\right).}
$$
These are ordinary topological Euler characteristics, rather than <orbifold> Euler characteristics. If the lifted surface is smooth, the equivalent <orbifold> formula is
$$
\chi(M)=|T|\chi_{\rm orb}(M_0),\qquad
\chi_{\rm orb}(M_0)=\chi(M_0)-\sum_j\left(1-\frac1{m_j}\right).
$$
For a <closed> oriented smooth surface, $\chi=2-2g$ defines its <genus>. These formulas also follow from the <Riemann-Hurwitz formula> by subtracting the ramification deficits $|T|-|T|/q_j$ over each base vertex.
For $U\le T$, the quotient map $M\to M_1=U\backslash M$ is an ordinary normal covering if and only if $U$ acts freely. All possible point stabilizers in this construction are conjugates of $\langle C_j\rangle$, so the concrete necessary and sufficient condition is
$$
\boxed{U\cap g\langle C_j\rangle g^{-1}=\{1\}\quad\text{for every }j\text{ and every }g\in T.}
$$
This is the <freeness criterion for a subgroup of a polygon-cover deck group>. Under it the deck <group> is $U$, the degree is $|U|$, and
$$
\boxed{\chi(M_1)=\frac{\chi(M)}{|U|}.}
$$
There is no requirement $U\triangleleft T$ for this map. That different condition characterizes normality of the intermediate cover $M_1\to M_0$. If fixed points are retained and one instead speaks of an <orbifold> covering, the quotient is regular as an <orbifold> cover, but ordinary <Euler characteristic> need not divide by $|U|$; <orbifold Euler characteristic> does.
For the low-genus examples take
$$
T=\operatorname{GL}(3,\mathbb F_2)=\operatorname{PSL}(3,2),\qquad |T|=(8-1)(8-2)(8-4)=168.
$$
Let $U_1$ be a nonzero-vector stabilizer and $U_2$ a plane stabilizer. Their actions have seven points, so $|U_1|=|U_2|=24$ and $[T:U_i]=7$. They are <almost conjugate subgroups>: the numbers of fixed nonzero vectors for a matrix $A$ and its dual action $A^{-T}$ are both $2^{\dim\ker(A-I)}-1$. Thus the two coset <permutation characters> agree, which is equivalent to <Gassmann equivalence>. They are not conjugate: a point stabilizer has a common fixed nonzero vector, whereas a plane stabilizer fixes no nonzero vector globally. In particular, since $7\nmid24$, neither <subgroup> contains any nonidentity element of a cyclic <subgroup> of order seven. All cone monodromies of exact order seven therefore act freely on the intermediate covers.
For <genus> three, start with a <hyperbolic triangle> whose three angles are $\pi/7$ and double it across one edge. The resulting quadrilateral $D$ has angles
$$
\boxed{\pi/7,\ 2\pi/7,\ \pi/7,\ 2\pi/7.}
$$
The side pairings produce three base vertex cycles, each of total angle $2\pi/7$, and an underlying sphere. Choose generators $A,B$ with $|A|=|B|=|AB|=7$; the three vertex words are $A,B,(AB)^{-1}$. Hence $s=2$, $r=3$ and all $q_j=7$. The full cover is smooth, and
$$
\chi(M)=168\left(-1+\frac37\right)=-96,\qquad
\chi(U_i\backslash M)=-96/24=-4,\qquad\boxed{g_i=3.}
$$
The <hyperbolic polygon area> of $D$ is $8\pi/7$, so the seven-sheeted surfaces have <area> $8\pi$, agreeing with Gauss-Bonnet. Their equal spectra follow from the Sunada theorem. For complete numerical generator data, one possible pair is
$$
A=\begin{pmatrix}0&0&1\\0&1&1\\1&1&0\end{pmatrix},\qquad
B=\begin{pmatrix}0&0&1\\1&0&0\\0&1&1\end{pmatrix}
\quad\text{over }\mathbb F_2.
$$
These generate all of $T$ and have the required orders. This construction establishes the requested <isospectrality>; it does not establish nonisometry. In the constant-curvature triangle example a lifted reflection can identify the point and plane covers, as in <reflection intertwining of triangle-cover coset actions>. Breaking that symmetry with a generic metric gives nonisometric genus-three surfaces, but then the metric need not have constant curvature. Those are distinct assertions.
For <genus> four, use the <cone-torus construction of genus-four Sunada surfaces>. Take a regular hyperbolic quadrilateral with four angles
$$
\boxed{\pi/14,\ \pi/14,\ \pi/14,\ \pi/14}
$$
and pair opposite sides in the usual <torus> pattern $a,b,a^{-1},b^{-1}$. All four vertices belong to one cycle, whose total angle is $2\pi/7$; the underlying base is a <torus> with one <cone point>. Choose $A,B$ of order four generating $T$, with commutator $[A,B]$ of order seven. For example,
$$
A=\begin{pmatrix}0&0&1\\0&1&0\\1&1&0\end{pmatrix},\qquad
B=\begin{pmatrix}0&0&1\\0&1&1\\1&0&0\end{pmatrix},\qquad
|A|=|B|=4,\quad |[A,B]|=7.
$$
Only the commutator is a cone monodromy, so the order-four side labels do not create cone points of order four. Here $s=2$, $r=1$, $q_1=7$, giving
$$
\chi(M)=168\left(-1+\frac17\right)=-144,\qquad
\chi(U_i\backslash M)=-144/24=-6,\qquad\boxed{g_i=4.}
$$
The quadrilateral has <area> $12\pi/7$, and each seven-sheeted smooth quotient has <area> $12\pi$. Again the order-seven cone stabilizer meets neither $U_i$ nontrivially, so the metrics are smooth and the Sunada theorem gives identical spectra. The required <group> and <subgroup> orders are therefore \b[168 and 24 in both constructions]; the genus-three side-generator orders are \b[7 and 7], and the displayed genus-four side-generator orders are \b[4 and 4], with cone-word order \b[7] in both cases.
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