= Solution
Write the <hyperbolic toral automorphism> as $T_A([v])=[Av]$ on $\mathbb T^2=\mathbb R^2/\mathbb Z^2$, where $A\in\mathrm{GL}(2,\mathbb Z)$. Its <determinant> is $\pm1$. The <eigenvalues> cannot be a nonreal conjugate pair: their product would have <absolute value> one, forcing each to have <absolute value> one, contrary to hyperbolicity. The same product condition and hyperbolicity imply two distinct real <eigenvalues>, labelled so that
$$
0<|\lambda_s|<1<|\lambda_u|,qquad |\lambda_s\lambda_u|=1.
$$
We will prove
$$
\boxed{h_{\mathrm{top}}(T_A)=\log|\lambda_u|.}
$$
This is the <entropy of a hyperbolic toral automorphism>; the <absolute value> is important when the expanding <eigenvalue> is negative.
Choose real <eigenvectors> $v_s,v_u$ and the adapted <norm> $\|a v_s+b v_u\|_*=\max(|a|,|b|)$. Its quotient <metric> on the <torus> is
$$
d([v],[w])=\min_{\ell\in\mathbb Z^2}\|v-w-\ell\|_*.
$$
It is a <compatible metric>, so part (a) allows its use. Let $\rho=\min_{0\ne\ell\in\mathbb Z^2}\|\ell\|_*>0$ and $D=\|A\|_*=|\lambda_u|$. Fix $\epsilon>0$ small enough that $2\epsilon<\rho$ and $(D+1)\epsilon<\rho$.
If $[v]$ lies in the <Bowen ball> $B_n(0,\epsilon)$, each $T_A^k[v]$ has a unique lift $z_k$ with $\|z_k\|_*<\epsilon$. Uniqueness follows because two such lifts differ by a lattice vector of <norm> less than $2\epsilon<\rho$. Moreover, $z_{k+1}-Az_k$ is a lattice vector, and
$$
\|z_{k+1}-Az_k\|_*<(D+1)\epsilon<\rho.
$$
It must be zero. Thus $z_k=A^kz_0$: there is no hidden lattice jump between successive lifted iterates.
Write $z_0=a v_s+b v_u$. The simultaneous bounds $\|A^kz_0\|_*<\epsilon$ for $0\leq k<n$ are exactly
$$
|a|<\epsilon,qquad |b|<\epsilon|\lambda_u|^{-(n-1)}.
$$
The stable coordinate is largest at time zero and the unstable coordinate at time $n-1$. Conversely, every vector in this rectangle satisfies all the <Bowen ball> bounds. The quotient is injective on this rectangle by $2\epsilon<\rho$. If $J=|\det(v_s,v_u)|$, its normalized <Haar measure> is therefore
$$
\nu(B_n(0,\epsilon))=4J\epsilon^2|\lambda_u|^{-(n-1)}.
$$
The quotient <metric> is translation invariant and $T_A$ is linear, so $d_n(x,y)=d_n(0,y-x)$; every <Bowen ball> has this same <measure>, independently of its centre.
Take an $\epsilon$-<separated set> of largest cardinality $s_n(d,\epsilon)$. It is maximal under inclusion, so its radius-$\epsilon$ <Bowen balls> cover the <torus>: a point outside all these balls could be added. The covering bound gives
$$
s_n(d,\epsilon)\geq\frac1{4J\epsilon^2}|\lambda_u|^{n-1}.
$$
Its radius-$\epsilon/2$ <Bowen balls> are disjoint, by the <triangle inequality> for $d_n$. Their measures sum to at most one, giving
$$
s_n(d,\epsilon)\leq\frac1{J\epsilon^2}|\lambda_u|^{n-1}.
$$
Taking logarithms, dividing by $n$, and letting $n\to\infty$ yields $h_d(T_A,\epsilon)=\log|\lambda_u|$ for every sufficiently small $\epsilon$. The limit as $\epsilon\downarrow0$ proves the answer. Contraction in the stable direction contributes no exponential growth; only expansion in the unstable direction does.
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