I am not an expert in this area, but I happen to see some references recently for equivalence 1. I am not sure about equivalence 2. The following are references from the texts. My comments are given as remarks.
References from [1]
Let $\Bbb T^d=\Bbb R^d/\Bbb Z^d$ denote the additive $d$-dimensional torus. This is a compact abelian group, and any continuous surjective endomorphism $T:\Bbb T^d\to \Bbb T^d$ is of the form
$$T(x)=T_A(x),$$
where $A$ is a non-singular $d\times d$ integer matrix, and $T_A$ is defined by
$$T_A \begin{bmatrix}x_1\\\vdots\\ x_d\end{bmatrix}=A\begin{bmatrix}x_1\\ \vdots\\ x_d\end{bmatrix} \pmod 1$$
Assume the transformation $T_A$ is an automorphism (that is, $\det(A)$ is $\pm 1$.) Then the transformation $T_A$ is said to be $\textit{expansive}$ if there is a constant $\delta>0$ with the property that for any pair $x\neq y\in \Bbb T^d$, there is an $n\in\Bbb Z$ with the property that
$$\rho(T_A^n x,T_A^n y)>\delta$$
$\textbf{Lemma 2.9}$ The invertible transformation $T_A$ is expansive if and only if no eigenvalue of $A$ has unit modulus.
This well-known result may be seen directly using the Jordan form of the complexified matrix (outlined in [2], chapter 8) or as part of the general theory of hyperbolic dynamical systems (see [3]). Eisenberg in [4] proved the analogous statement for matrices acting on topological vector spaces (which for real vector spaces implies the toral result) over any non-discrete topological field.
Remark: Here the conditions $\det(A)=\pm 1$ and eigenvalues of $A$ are not unit modulus will correspond to your stated definition of hyperbolic. The text only gives proof for dimension 2, although it does provide references for the general case. Also, you can download [4] freely.
References from [2] (page 143)
Let $A$ be an automorphism of the $n$-torus, and $[A]$ the corresponding matrix. Then $[A]$ is expansive iff $[A]$ has no eigenvalues of modulus 1.
Proof:
Let $d$ be the metric given by
$$d(\{x_n\},\{y_n\})=\sum_{n=-\infty}^\infty \frac{|x_n-y_n|}{2^{|n|}}.$$
Suppose $\{x_n\}\neq \{y_n\}$. Then for some $n_0,x_{n_0}\neq y_{n_0}$ and
\begin{align*}
d(T^{n_0}\{x_n\},T^{n_0}\{y_n\}) &=\sum_{n=\infty}^\infty \frac{1}{2^{|n|}} |x_{n+n_0} - y_{n+n_0}|\\&\geq |x_{n_0}-y_{n_0}|\\&\geq 1.\end{align*}
Thus 1 is an expansive constant.
Remark: This seems to answer equivalence 1 exactly, where you set $\epsilon = 1$.
Remark While [3] also appears to contain all the information for equivalence 1, I was unable to pull out a single proposition/theorem proving it as the hyperbolic toral automorphisms are considered as examples there.
References
[1] G. Everest and T. Ward, Heights of Polynomials and Entropy in Algebraic Dynamics, Universitext, Springer, 1999
[2] P. Walters, An Introduction to Ergodic Theory, Springer, New York 1982.
[3] A. Katok and B. Hasselblatt, Introduction to the Modern Theory of Dynamical Systems, Cambridge University Press, Cambridge, 1995.
[4] M. Eisenberg, Expansive automorphisms of finite-dimensional spaces, Fundamenta Math. LIX (1966), 307-312
Since $f$ is a continuous map on the compact metric space $\mathbb{T}^2$, one form of the variational principle says that
$$
h_{top}(f) = \sup_{\mu \in \mathscr{M}^f_e(\mathbb{T}^2)} h_\mu(f)
$$
where $\mathscr{M}^f_e(\mathbb{T}^2)$ is the set of all ergodic $f$-invariant probability measures on $\mathbb{T}^2$. So it is enough to show that $h_\mu(f) = 0$ for any ergodic $f$-invariant measure $\mu$.
A key observation is the following. If $L_y = \{(x,y): x \in \mathbb{T}\}$ is the horizontal line in $\mathbb{T}^2$ at height $y$, then each $L_y$ is an $f$-invariant set. Therefore if $\mu$ is any ergodic measure, there is a unique $y$ such that $\mu$ is supported on $L_y$. Now when restricted to the line $L_y$, the action of $f$ is clearly isomorphic to the circle rotation $x \mapsto x+y$ on $\mathbb{T}$. Since $\mu$ is supported on a line, it is naturally identified with a measure on $\mathbb{T}$, and so our system $(\mathbb{T}^2, f, \mu)$ is naturally isomorphic to the circle rotation $(\mathbb{T}, x \mapsto x+y, \mu)$.
It's well known that 1-dimensional circle rotations have zero topological entropy, so in particular any invariant measure for a circle rotation has zero measure entropy as well. Therefore $h_\mu(f) = 0$. Since this holds for any ergodic $f$-invariant measure $\mu$ we are done.
Best Answer
Your question leads to the following fact: If $L \subset \mathbb R^2$ is a lattice spanned by integer vectors, then the number of integer points in $\mathbb R^2 / L$ is the covolume of $L$ (area of a fundamental parallellogram for $L$).
This may be deduced either from
So you don't need the matrix $A$ to be hyperbolic, you also don't need $|\det A| = 1$; all you need is that $A - I$ is invertible, which is the case if $A$ is hyperbolic.