= Solution
For an open cover $X=A\cup B$, the <Mayer–Vietoris theorem> gives the long exact sequence
$$
\cdots\to H_j(A\cap B)\xrightarrow{(i_*,-j_*)}H_j(A)\oplus H_j(B)\to H_j(X)\to H_{j-1}(A\cap B)\to\cdots.
$$
The same sequence holds for a union of simplicial subcomplexes. All <homology groups> here have <integer> coefficients.
An explicit <triangulation> of $X_q$ is as follows. Take vertices $b_0,b_1,b_2$ forming a target circle, a circle of $3q$ vertices $a_0,\ldots,a_{3q-1}$, and a cone <vertex> $c$. With subscripts on $a$ interpreted modulo $3q$ and on $b$ modulo three, include triangles
$$
[a_j,a_{j+1},b_{j+1}],\qquad[a_j,b_{j+1},b_j],\qquad[c,a_j,a_{j+1}]
$$
for every $j$, together with all their faces. The first two types triangulate the <mapping cylinder> of the degree-$q$ covering $a_j\mapsto b_j$; the last type fills the inner circle by a disc. A filled disc with an attached annular collar is still a disc, and identifying the outer collar boundary by this covering gives exactly the source quotient. Distinct inner vertices ensure this is a genuine <simplicial complex>, not just a one-cell presentation.
Let $A$ be the <mapping cylinder> and $B$ the coned disc. Then $A$ retracts to the target circle, $B$ is contractible and $A\cap B$ is the inner circle. Its inclusion into $A$ induces multiplication by $q$ on $H_1$. The exact sequence therefore reduces to
$$
0\to H_2(X_q)\to\mathbb Z\xrightarrow{q}\mathbb Z\to H_1(X_q)\to0.
$$
Connectedness and the absence of higher-dimensional simplices give
$$
\boxed{H_0(X_q)=\mathbb Z,\qquad H_1(X_q)=\mathbb Z/q\mathbb Z,\qquad H_j(X_q)=0\quad(j\ge2).}
$$
This is a two-dimensional <Moore space> with torsion in degree one.
The <suspension> of a connected complex is the union of two cones whose intersection retracts to the original complex. Reduced Mayer-Vietoris gives
$$
\boxed{\widetilde H_j(SK)\cong\widetilde H_{j-1}(K)\quad(j\ge1),\qquad H_0(SK)=\mathbb Z.}
$$
In particular its $H_1$ vanishes. For the wedge of two connected polyhedra, choose neighborhoods retracting onto the two summands with contractible intersection at the wedge point. Then
$$
\boxed{\widetilde H_j(X\vee Y)\cong\widetilde H_j(X)\oplus\widetilde H_j(Y).}
$$
Decompose each prescribed $G_i$ into cyclic summands. Realize each infinite cyclic summand in degree $i$ by $S^i$, and each finite nontrivial summand $\mathbb Z/q$ by $S^{i-1}X_q$, meaning $i-1$ successive <suspensions>. Take their finite wedge, or a point if there are no summands. The <suspension> formula shifts the torsion to degree $i$ and creates no other positive-degree homology, while the wedge formula adds summands. Thus the resulting connected <polyhedron> satisfies \b[$\boxed{H_0(X)=\mathbb Z,\ H_i(X)=G_i\ (1\le i\le n),\ H_i(X)=0\ (i>n)}$]. The finite <triangulations> above, their simplicial <suspensions> and wedges provide an actual <polyhedron>.
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