1) Yes. $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Q}$ is the localization $(\mathbb{Z}\setminus \{0\})^{-1} \mathbb{Z}_p$. Thus, elements have the form $a/b$ with $a \in \mathbb{Z}_p$ and $b \in \mathbb{Z} \setminus \{0\}$. Clearly this is a subring of $\mathbb{Q}_p$. In order to show that it is the whole of $\mathbb{Q}_p$, it suffices to prove that it contains all $1/u$ for $u \in \mathbb{Z}_p \setminus \{0\}$, i.e. that there are $a,b$ as above satisfying $b=ua$, i.e. that $u$ divides some positive integer in $\mathbb{Z}_p$. But actually $u$ is associated to some positive integer, namely to $p^n$ where $n$ is the $p$-adic valuation of $u$.
Actually this shows that already the localization at the element $p$ gives $\mathbb{Q}_p$. More generally, if $R$ is a DVR with uniformizer $\pi$, then $R_{\pi}=Q(R)$.
2) Yes, If $n$ is any positive integer, you can define $\mathbb{Z}_n := \varprojlim_k~ \mathbb{Z}/n^k$, the $n$-adic completion of $\mathbb{Z}$. The Chinese Remainder Theorem gives $\mathbb{Z}_{nm} \cong \mathbb{Z}_n \times \mathbb{Z}_m$ for coprime $n,m$, and we have $\mathbb{Z}_{n^v}=\mathbb{Z}_{n}$ for $v>0$ since limits of cofinal subsystems agree. Thus, if $n = p_1^{v_1} \cdot \dotsc \cdot p_n^{v_n}$ is the prime decomposition of $n$ with $v_i > 0$, then $\mathbb{Z}_n \cong \mathbb{Z}_{p_1} \times \dotsc \times \mathbb{Z}_{p_n}$. In principle one gets nothing new.
3) I don't think that there is a nice description of $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_q$. The tensor product behaves well for finite products and all colimits, but $\mathbb{Z}_p$ is an infinite projective limit. So you should better consider $\mathbb{Z}_p \widehat{\otimes} \mathbb{Z}_q$, some completed tensor product, having in mind that the $p$-adics form a (very nice) topological ring. I suspect that this is a ring which has not been considered in the literature, but I am not sure ...
Perhaps someone else can add a reference on the tensor product of topological rings, because I could only find this for topological $\mathbb{C}$-algebras.
Usually "rational integer" and "rational prime" is terminology thrown around in algebraic number theory to mean integer of $\mathbb{Q}$ and prime of (the ring of integers of) $\mathbb{Q}$, i.e. elements of $\mathbb{Z}$ or primes in $\mathbb{Z}$. This terminology is used to differentiate between the term algebraic integer, which will often be said without the algebraic preceding it, and the term prime which might refer to a prime ideal in the ring of integers of a number field.
Best Answer
You might look at (1). For representing $a/b \in \Bbb{Q}$ in $\Bbb{Q}_p$, the basic method (for rationals having $p$-adic absolute value $1$) is to solve the congruence $p^k \cong 1 \pmod{b}$ for $k$, so that $p^k - 1 = b \cdot c$, for some integer $c$. Then $$ \frac{a \cdot c}{b \cdot c} = \frac{ac}{p^k - 1} = \frac{-ac}{1 - p^k} \text{,} $$ which, suggests a geometric series in powers of $p^k$. Now express the integer $-ac$ in base $p$ and this is the sequence of digits repeating in the $p$-adic representation.
For $a/b > 0$, it is sometimes easier to compute the representation of $-a/b$, then negate the $p$-adic result.
For rationals having $p$-adic absolute value $\neq 1$, there is a preperiodic part of the expansions and a periodic part. The periodic part is obtained similarly to the above. The preperiodic part is a bit more work, as described in the linked article.
(1) Conrad, K. "The $p$-adic expansion of rational numbers", https://kconrad.math.uconn.edu/blurbs/gradnumthy/rationalsinQp.pdf .