Question: Is there a ring homomorphism between $\mathbb{Z}$ into $\mathbb{Z} \times \mathbb{Z}$?

Also for any rings $R\_{1}$ and $R\_{2}$ does there exist a ring homomorphism $\phi$ : $R\_{1} \rightarrow$ $R\_{1} \times R\_{2}$?

Note: I am allowing 1 in both rings. So any homomorphism must map the identity on one ring to the identity on the other ring.

Answer: 
Suppose $R$, $S$ and $T$ are rings. Giving a ring homomorphism $f\colon R\to S\times T$ is the same as giving homomorphisms $g\colon R\to S$ and $h\colon R\to T$.

Let's see why. First, the projection maps $p\colon S\times T\to S$ and $q\colon S\times T\to T$ are ring homomorphisms, so if we are given $f\colon R\to S\times T$, we get $p\circ f\colon R\to S$ and $q\circ f\colon R\to T$.

Suppose instead we are given $g\colon R\to S$ and $h\colon R\to T$. Define
$$
f(x)=(g(x),h(x))
$$
It's easy to see that $f$ is a ring homomorphism and that $g=p\circ f$, $h=q\circ f$.

In general terms this is the statement that $S\times T$ is the *product* in the category of rings.

This answers your second question: in order to have a ring homomorphism $R\to R\times T$ you need a ring homomorphism $R\to T$ (and use, for instance, the identity as the homomorphism $R\to R$). So, for a counterexample find rings $R$ and $T$ such that there is no ring homomorphism $R\to T$; take $R$ to have characteristic $2$ and $T$ to have characteristic $3$, for instance.

In the case of $\mathbb{Z}$, it's a standard result that, for any ring $R$, there is a unique ring homomorphism $\mathbb{Z}\to R$ (when unital rings are concerned and homomorphisms are required to preserve the identity element).
As for the second question:

There is no homomorphism $\mathbb Z\_5\rightarrow \mathbb Z\_5\times \mathbb Z$.

This is because the image must be isomorphic to a quotient of $\mathbb Z\_5$, and is therefore finite. On the other hand it contains $n(1,1)$ for every $n$, this gives us an infinite number of elements.