I want to emphasize that $\mathbb Z_p$ is not just $\mathbb F_p[[X]]$ in disguise, though the two rings share many properties. For example, in the $3$-adics one has
\[
(2 \cdot 1) + (2 \cdot 1) = 1 \cdot 1 + 1 \cdot 3 \neq 1 \cdot 1.
\]
I know three ways of constructing $\mathbb Z_p$ and they're all pretty useful. It sounds like you might enjoy the following description:
\[
\mathbb Z_p = \mathbb Z[[X]]/(X - p).
\]
This makes it clear that you can add and multiply elements of $\mathbb Z_p$ just like power series with coefficients in $\mathbb Z$. The twist is that you can always exchange $pX^n$ for $X^{n + 1}$. This is the “carrying over” that Thomas mentions in his helpful series of comments.
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.
Best Answer
You want a map from ${\bf Z}[[x]]$ to that ring of coefficient-restricted power series, with kernel generated by $p-x$. Here's a start on constructing such.
Any positive integer can be written in base $p$ as $a=a_0+a_1p+\cdots+a_rp^r$ with $0\le a_i\le p-1$ for each $i$. This gives you a map $\phi$ from positive integers to (coefficient-restricted) polynomials by $\phi(a)=a_0+a_1x+\cdots+a_rx^r$. Now all you have to do is extend the domain from the positive integers to ${\bf Z}[[x]]$.
Start with the negative integers. In fact, start with $-1$; $$-1=(p-1)+(p-1)p+(p-1)p^2+\cdots$$ so $$\phi(-1)=(p-1)+(p-1)x+(p-1)x^2+\cdots$$ Now you can get $\phi(n)$ for any negative integer $n$ as a coefficient-restricted power series - details left to the reader.
The more complicated problem is what to do after you've applied $\phi$ to each coefficient of an element of ${\bf Z}[[x]]$ and because of the interaction between coefficients you still have some (perhaps infinitely many) coefficients outside the desired range. Well, apply $\phi$ again, and again, and again. After $k$ applications, at least the first $k$ coefficients will be OK, and they will stay OK forever after, so in the limit, you have your isomorphism.
There's probably a way of stating this in finite terms, but I'm not seeing it right now.