Equivalence relation: cardinality of quotient set 
At $A=\mathbb{Z}^{\mathbb{N}}$ we define equivalence relation $\equiv$ by:
$$f\equiv g \iff \forall n\in \mathbb{N} ((f(2n)=g(2n)) \wedge(f(n)\cdot g(n)> 0  \ \vee f(n)=g(n)=0)) $$
a) Find the cardinality of the quotient set
b) Find the equivalence class for the constant function with value $0$
c) Decide if all equivalence classes have the same cardinality

My try:
To a) I think it's $|\mathbb{R}|$.
b) It's just $1$?
c) No since $f(n)=0$ has $1$ and if we define for $f(n)=n$, $g(2n)=2n$ and $g(2n+1)=k$ then this class is $|\mathbb{N}|$
 A: Your answers are correct, but particularly part a) could do with some elaboration. Also, your second equivalence class in c) has cardinality $|\Bbb R|$, can you see why?
The simplest way to establish a) is by estimation. On the one hand, we have $|\Bbb Z^{\Bbb N}| = |\Bbb R|$, so that $A / \equiv$ has at most this cardinality.
On the other hand, we exhibit the following set of inequivalent functions:
$$\forall g \in \Bbb Z^{\Bbb N}: f_g(n) = \begin{cases}0 &:\text{$n$ odd} \\ g(m)&:\text{$n = 2m$}\end{cases}$$
because we easily see $f_g \equiv f_h \iff g = h$. Therefore, there are also at least $|\Bbb R|$ equivalence classes, establishing the result.

Of separate interest is the determination of all equivalence classes. It is not too hard to see that on odd $n$, two functions $f$ and $g$ being equivalent amounts to:
$$\operatorname{sgn}(f(2n+1)) = \operatorname{sgn}(g(2n+1))$$
for all $n$, where $\operatorname{sgn}(n)$ is $0, 1$ or $-1$ according to whether $n$ is zero, positive or negative. It follows that we can describe all equivalence classes by the functions $f_{g,s}$ (where $g \in \Bbb Z^{\Bbb N}$ and $s \in \{0,1,-1\}^{\Bbb N}$) defined as:
$$f_{g,s}(n) = \begin{cases}g(m) &: n = 2m\\s(m) &: n = 2m+1\end{cases}$$
