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.
Here is a general criterion. (All tensor products in this answer are over $R$.)
Theorem: Let $R$ be a ring and let $M$ be a right $R$-module. Then the following are equivalent.
The functor $M\otimes -$ preserves products: that is, for every family $(A_i)$ of left $R$-modules, the canonical map $M\otimes\prod A_i\to \prod M\otimes A_i$ is an isomorphism.
$M$ is finitely presented.
Proof: First, suppose $M$ has a finite presentation $$R^m\to R^n\to M\to 0$$ and let $(A_i)$ be any family of left $R$-modules. We then get a commutative diagram
$$\require{AMScd}
\begin{CD}
R^m\otimes \prod A_i @>>> R^n\otimes \prod A_i @>>> M\otimes \prod A_i @>>> 0\\
@VV{}V @VV{}V @VV{}V \\
\prod R^m\otimes A_i @>>> \prod R^n\otimes A_i @>>> \prod M\otimes A_i @>>> 0
\end{CD}$$ whose rows are exact. Now note that $R^m\otimes \prod A_i\cong (\prod A_i)^m$ and $\prod R^m\otimes A_i\cong \prod A_i^m$, and our vertical map between them is easily seen to be the canonical isomorphism which interchanges the products. So the left vertical map is an isomorphism and similarly so is the middle vertical map. By the five lemma, it follows that the right vertical map is an isomorphism, and thus $M\otimes -$ preserves products.
Conversely, suppose $M\otimes-$ preserves products. In particular, then, the canonical map $$\varphi: M\otimes R^M\to M^M$$ is an isomorphism. Considering the identity map $id:M\to M$ as an element of the product $M^M$, we have $$id=\varphi(\sum m_i \otimes f_i)$$ for some finite collection of elements $m_i\in M$ and $f_i:M\to R$. Evaluating both sides of this equation at an element $m\in M$ we find $$m=\sum m_if_i(m).$$ Thus every element of $M$ is a linear combination of the elements $m_i$, so $M$ is finitely generated.
Now let $$0\to K\to F\to M\to 0$$ be a presentation of $M$ with $F$ a finitely generated free module. To conclude $M$ is finitely presented, we must show that $K$ is finitely generated. Now let $(A_i)$ be any family of flat left $R$-modules and consider the commutative diagram
$$\require{AMScd}
\begin{CD}
K\otimes \prod A_i @>>> F\otimes \prod A_i @>>> M\otimes \prod A_i @>>> 0\\
@VV{}V @VV{}V @VV{}V \\
\prod K\otimes A_i @>{\alpha}>> \prod F\otimes A_i @>>> \prod M\otimes A_i @>>> 0
\end{CD}$$
which again has exact rows. The right vertical map is an isomorphism by hypothesis and the middle vertical map is an isomorphism since $F$ is finitely generated and free. Moreover, the map $\alpha$ is injective since the $A_i$ are flat. A simple diagram chase now shows that the left vertical map is surjective.
Thus the canonical map $K\otimes \prod A_i\to \prod K\otimes A_i$ is surjective for any family of flat modules $(A_i)$. In particular, the canonical map $K\otimes R^K\to K^K$ is surjective, which we have seen above implies that $K$ is finitely generated.
By similar arguments, you can show that the canonical map $M\otimes \prod A_i\to\prod M\otimes A_i$ is always a surjection iff $M$ is finitely generated. Indeed, the forward direction is already contained in the forward direction of the proof above, and the reverse direction is similar to the proof of the reverse direction above.
Best Answer
You need to know that $\mathcal{O}$ is a finite free module. In general, tensoring with a module doesn't infinite products unless the module is finitely presented; see this answer for a proof.
In this case we can be more explicit. There is a natural map
$$\mathcal{O} \otimes \widehat{\mathbb{Z}} \to \prod_p \mathcal{O}_p$$
and to show that it's bijective we describe the underlying map of abelian groups as
$$\mathcal{O} \otimes \widehat{\mathbb{Z}} \cong \mathbb{Z}^2 \otimes \widehat{\mathbb{Z}} \cong \widehat{\mathbb{Z}}^2 \cong \prod_p \mathbb{Z}_p^2 \cong \prod_p \mathbb{Z}^2 \otimes \mathbb{Z}_p \cong \prod_p \mathcal{O}_p.$$
This is a composition of isomorphisms (of abelian groups) hence is an isomorphism (of rings).