Is $\epsilon^2/\epsilon^2=1$ or $0/0$? Is it possible in the system of dual numbers ($a+\epsilon b$; $\epsilon^2=0$) to calculate $\epsilon/\epsilon =1$? How then does one deal with $\epsilon^2/\epsilon^2=1$ versus $\epsilon^2/\epsilon^2=0/0$?
The same question for infitesimal calculus using hyperreal numbers where: $\epsilon \neq 0$ but $\epsilon^2=0$?
I probably did not use the correct formulation w.r.t. hyperreal numbers. I meant the axiom (?) in smooth infinitesimal analysis where it is assumed: $\epsilon \neq 0$ but $\epsilon^2=0$.
I am not quite sure how this analysis is related to nonstandard-analysis and hypercomplex numbers. I came across this topic in the book: A Primer of infinitesimal analysis (John L. Bell).
 A: In the dual numbers, ${\mathbb R}[\epsilon]$ ($={\mathbb R}[X]/(X^2)$), $\epsilon$ is not invertible, so the expression $\epsilon / \epsilon$ ($= \epsilon \epsilon^{-1})$ is undefined.
In hyperreals, as Asaf Karagila mentions in the comments, $\epsilon^2 \neq 0$. There you do have $\epsilon / \epsilon = \epsilon^2 / \epsilon^2 = 1$ (as the hyperreals are a field and $\epsilon$ is a non-zero element).
I had a very quick look at the book by Bell. That's not only using infinitesimals, but also a different kind of logic (no law of excluded middle!). That's not for the faint-of-heart :-): for a given $x$, the statement "$x = 0 \lor x \neq 0$" is not necessarily true in that setting.
A: "Multiplication by $\epsilon$" gives an isomorphism $\mathbf{R} \to \epsilon \mathbf{R}[\epsilon]$ (i.e. the ideal of multiples of $\epsilon$ in the dual numbers). Thus it's not terribly abusive to write "division by $\epsilon$" as the inverse of this isomorphism. If you do so, then $\epsilon / \epsilon$ is the real number 1 (and not the dual number $1$).
More generally, in any one-dimensional vector space over $\mathbf{R}$, it is not terrible to define division of vectors to be the unique real number that satisfies the equation $(v/w)w = v$.
Generalizing in another direction, if $I$ is a principal ideal of a ring $R$ generated by $t$, then "multiplication by $t$" gives a surjective $R$-module homomorphism $\mu_t : R \mapsto I : r \mapsto tr$ (this is not a ring homomorphism!). Correspondingly, it is not terrible to define "division by $t$" on $I$ to be the inverse isomorphism $I \mapsto R / \ker \mu_t$ that maps $tr \mapsto r$.
e.g. if I write $f(x) + O(x^n)$ to be the equivalence class of $f(x)$ modulo $(x^n)$ in the ring $\mathbf{R}[x]$, then we would have the appealing fact
$$ \frac{xf(x) + O(x^n)}{x} = f(x) + O(x^{n-1}) $$
A: As far as smooth infinitesimal analysis (SIA) is concerned, it is not correct that it assumes that $\epsilon \neq 0$.  It cannot be proven that $\epsilon \neq 0$.  What is true is that infinitesimals in SIA satisfy $\neg\neg(\epsilon = 0)$.
