= Solution
Take two <solid tori> $V_1,V_2$, each with a meridian-longitude <basis> $(\mu_i,\lambda_i)$ on its boundary. Use opposite boundary-basis orientations on the two sides, so that a determinant-one coefficient matrix describes an <orientation>-reversing gluing map. For $A\in SL(2,\mathbb Z)$ choose a torus <homeomorphism> $f:\partial V_2\to\partial V_1$ inducing
$$
f_*\mu_2=-q\mu_1+p\lambda_1,\qquad
f_*\lambda_2=s\mu_1+r\lambda_1,
$$
and define $M_A=V_1\cup_fV_2$. The linear map on $\mathbb R^2/\mathbb Z^2$ realizes the matrix. Torus <homeomorphisms> with the same action on <first homology> are isotopic, and an isotopy extends over a boundary collar, so the resulting <three-manifold> depends only on $A$. Specifying opposite boundary conventions is necessary: gluing two identically oriented boundary bases by a determinant-one map would not describe the usual oriented gluing.
We now identify this gluing directly in the quotient from part (a). Split $S^3$ into the two invariant <solid tori> $W_1=\{|z_2|\le1/\sqrt2\}$ and $W_2=\{|z_1|\le1/\sqrt2\}$. On $W_1$ write $z_1=\sqrt{1-|w|^2}e^{2\pi ix}$, $z_2=w$. The quotient coordinates
$$
(u,v)=(e^{2\pi ipx},e^{-2\pi iqx}w)
$$
identify $W_1/\langle g\rangle$ with $S^1\times D^2$. They are invariant under $g$; choosing a $p$th root of $u$ reconstructs $(x,w)$, and different choices are exactly the same cyclic orbit. Thus this is a genuine solid-torus coordinate model.
On the common boundary, measure arguments in turns as $(x,y)$. Its quotient lattice is
$$
\mathcal L=\mathbb Z^2+\mathbb Z(1/p,q/p).
$$
The meridian and longitude of the first quotient torus can be chosen as $\mu_1=(0,1)$ and $\lambda_1=(1/p,q/p)$. They form a lattice <basis>, since $(1,0)=p\lambda_1-q\mu_1$. The meridian of the second torus is precisely that $(1,0)$ loop. Consequently
$$
\boxed{\mu_2=-q\mu_1+p\lambda_1.}
$$
The determinant condition is $qr+ps=-1$. Put $k=-r$, so $kq\equiv1\pmod p$. On $W_2$, the analogous quotient coordinates use $e^{2\pi ipy}$ and $e^{-2\pi iky}z_1$, and a longitude is $(k/p,1/p)$. Choose its negative for the second boundary-basis convention. Then
$$
\lambda_2=(-k/p,-1/p)=s\mu_1+r\lambda_1,
$$
since $ps+qr=-1$. Thus the two quotient tori are glued by exactly $A$. This proves the <genus-one gluing model of a lens space> and
$$
\boxed{M_A\cong L(p,q).}
$$
Different choices of $r,s$ satisfying the determinant condition differ by $(s,r)\mapsto(s,r)+j(-q,p)$. This changes $f$ by a meridional <Dehn twist> on $V_2$, which extends over $V_2$: in solid-torus coordinates it is $(z,e^{i\theta})\mapsto(e^{ij\theta}z,e^{i\theta})$. Therefore there is no hidden dependence on that choice of longitude.
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