Describe all ring homomorphisms

$1$: All ring homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}$

Let $f: \mathbb{Z} \to \mathbb{Z} $ be a ring homomorphism. Note that for $n \in \mathbb{Z}$, $f(n) = nf(1)$. Thus $f$ is completely determined by its value on $1$. Since $1$ is an idempotent in $\mathbb{Z} $ (i.e. $1^2 = 1$), then $f(1)$ is again idempotent. Now we need to determine all of the idempotents of $\mathbb{Z} $. To this end, take $x\in \mathbb{Z}$ such that $x^2 = x$. Thus $x^2 − x = x(x − 1) = 0$. Since $\mathbb{Z} $ is an integral domain, we deduce that either $x = 0$ or $x = 1$. Thus the complete list of idempotents of $\mathbb{Z} $ are $0$ and $1$. Thus $f(1)$ being idempotent implies that either $f(1) = 0$ or $f(1) = 1$. In the first case, $f(n) = 0$ for all $n$ and in the second case $f(n) = n$ for all $n$. Thus, the only ring homomorphisms from $\mathbb{Z} $ to $\mathbb{Z} $ are the zero map and the identity map.

$2$: All ring homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}\times \mathbb{Z}$

Let $f$ be such a ring homomorphism. Suppose that $f(1) = (a, b) $, with $a, b \in \mathbb{Z} $. Since $f$ is a ring homomorphism it follows that In particular, $f(m) = f(m · 1) = m · f(1) = m(a, b)$ (follows from the additivity of $f$). On the other hand, $f$ preserves multiplication. That is, $f(mn) = f(m)f(n)$. Thus, we have

$mn(a, b) = m(a, b) · n(a, b)$ iff

$mn(a, b) = mn(a^2, b^2)$ iff

$(a, b) = (a^2, b^2)$ if $mn\neq 0$.

This last inequality only holds if $a = 0, 1$ and $b = 0, 1$. It follows that there are four ring homomorphisms which are given by

$f_1(1) = (0, 0)$, $f_2(1) = (1, 0)$, $f_3(1) = (0, 1)$, $f_4(1) = (1, 1)$.

More explicitly, these are $f_1(m) = (0, 0)$, $f_2(m) = (m, 0)$, $f_3(m) = (0,m)$, $f_4 (m) = (m,m)$.

$3$: All ring homomorphisms from $\mathbb{Z}\times \mathbb{Z}$ to $\mathbb{Z}$

Since $\mathbb{Z}\times \mathbb{Z}$ is generated by $(1, 0)$ and $(0, 1)$, it suffices to find to find $f(1, 0)$ and $f(0, 1)$

Leaving this for you.

$4$: All ring homomorphisms from $\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}$ to $\mathbb{Z}$

Hint. Find $f(1, 0, 0)$ $f(0, 1, 0)$ and $f(0, 0,1)$

There is only one ring homomorphism from $\mathbb{Z}$ to any ring $S$ (assuming that ring homomorphisms preserve $1$).

This homomorphism is uniquely determined because a ring homomorphism $f\colon R\to S$ has the property that

$$f(nx) = n f(x)$$

for all $x\in R$ and $n\in\mathbb{Z}$ (easy induction). Since in $\mathbb{Z}$ we have $n=n\cdot1$, for $f\colon\mathbb{Z}\to S$ we must have $f(n)=nf(1)=n\cdot1$. The assignment

$$f(n)=n\cdot1$$

indeed defines a ring homomorphism from $\mathbb{Z}$ to any ring $S$. Its kernel is an ideal of the form $k\mathbb{Z}$ $(k\ge0)$ and $k$ is the characteristic of $S$.

This answers your first two questions.

Let now $f\colon \mathbb{Z}\times\mathbb{Z}\to \mathbb{Z}$ be a ring homomorphism. Since

$$(1,0)(0,1)=(0,0)$$

we have that $f(1,0)f(0,1)=0$, so either $f(1,0)=0$ or $f(0,1)=0$. Not both, because $(1,1)=(1,0)+(0,1)$ is the identity on $\mathbb{Z}\times\mathbb{Z}$ and we want $f(1,1)=1$. Moreover we see that

$$f(0,1)=1-f(1,0).$$

If $f(1,0)=1$ and $f(0,1)=0$, then $f$ is completely determined: we must have

$$f(m,n)=f\bigl(m(1,0)+n(0,1))=mf(1,0)+nf(0,1)=m$$

It's easy to check that this indeed defines a ring homomorphism. Similarly

$$g(m,n)=n$$

defines another ring homomorphism and the list is complete.

For $\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$ the reasoning is similar. Consider

$$(1,0,0)+(0,1,0)+(0,0,1)=(1,1,1)$$

and the fact that two among $f(1,0,0)$, $f(0,1,0)$ and $f(0,0,1)$ must be $0$.