Solution:

Setting $y=0$ in the condition (ii), we get
$$(f(x)-f(0)) f(x)=0$$
for all $x$ (since $f(0) \neq 0$). Thus either $f(x)=0$ or $f(x)=f(0)$, for all $x \in \mathbb{Z}$. Now taking $x=y=0$ in (i), we see that $f(0)+f(0)^2=2 f(0)$. This shows that $f(0)=0$ or $f(0)=1$. Since $f(0) \neq 0$, we must have $f(0)=1$. We conclude that
$$\text{either } f(x)=0 \text{ or } f(x)=1 \text{ for each } x \in \mathbb{Z}$$
This shows that the set of all possible values of $f(x)$ is $\{0,1\}$. This completes (a).

Let $S=\{n \in \mathbb{Z} \mid f(n) \neq 0\}$. Hence we must have $S=\{n \in \mathbb{Z} \mid f(n)=1\}$ by (a). Since $f(1)=0$, $1$ is not in $S$. And $f(0)=1$ implies that $0 \in S$. Take any $x \in \mathbb{Z}$ and $y \in S$. Using (ii), we get
$$f(xy)+f(x)=f(x)+1$$
This shows that $xy \in S$. If $x \in \mathbb{Z}$ and $y \in \mathbb{Z}$ are such that $xy \in S$, then (ii) gives
$$1+f(x)f(y)=f(x)+f(y)$$
Thus $(f(x)-1)(f(y)-1)=0$. It follows that $f(x)=1$ or $f(y)=1$; i.e., either $x \in S$ or $y \in S$. We also observe from (ii) that $x \in S$ and $y \in S$ implies that $f(x-y)=1$ so that $x-y \in S$. Thus $S$ has the properties:

(A) $x \in \mathbb{Z}$ and $y \in S$ implies $xy \in S$;

(B) $x, y \in \mathbb{Z}$ and $xy \in S$ implies $x \in S$ or $y \in S$;

(C) $x, y \in S$ implies $x-y \in S$.

Now we know that $f(10) \neq 0$ and $f(2)=0$. Hence $f(10)=1$ and $10 \in S$; and $2 \notin S$. Writing $10=2 \times 5$ and using (B), we conclude that $5 \in S$ and $f(5)=1$. Hence $f(5k)=1$ for all $k \in \mathbb{Z}$ by (A).

Suppose $f(5k+l)=1$ for some $l$, $1 \leq l \leq 4$. Then $5k+l \in S$. Choose $u \in \mathbb{Z}$ such that $lu \equiv 1 \pmod{5}$. We have $(5k+l)u \in S$ by (A). Moreover, $lu=1+5m$ for some $m \in \mathbb{Z}$ and
$$(5k+l)u=5ku+lu=5ku+5m+1=5(ku+m)+1$$
This shows that $5(ku+m)+1 \in S$. However, we know that $5(ku+m) \in S$. By (C), $1 \in S$ which is a contradiction. We conclude that $5k+l \notin S$ for any $l$, $1 \leq l \leq 4$. Thus
$$S=\{5k \mid k \in \mathbb{Z}\}$$