Perhaps you should separate this into at least two questions, as your first two questions are "local" and your last three are "global".
As to the local ones:
This response is more an idea than a complete answer. But it seems to me that the Krasner-Serre mass formula should tell you whatever you want to know about the number of extensions of a $p$-adic field of given degree. (Note that it comes down to counting totally ramified extensions, since any other kind is much easier to count.) Casting about just now for a good reference, I looked through my own course notes on local fields and was severely disappointed: I say too little and what I do say is riddled with typos. But this paper of Pauli and Roblot seems to be, among other things, a very thorough survey of these $p$-adic mass formulas. In particular it contains references to the original papers of Krasner (1966) and Serre (1978).
I haven't looked at the details myself, but surely (meaning, of course, that I am not completely sure!) this mass formula will answer your first question. It also seems to have a good chance to answer your second question, possibly along with some inclusion-exclusion/Mobius inversion arguments.
As to the global ones:
3) Yes! This is a famous theorem of Hermite. Look in a good algebraic number theory book, e.g. one written by Neukirch.
4) I don't, no, off the top of my head, but others surely do. Stay tuned...
5) I guess I don't see why this should be true, but I'll have to think more about it.
As above, asking fewer questions at a time will probably elicit more detailed answers.
I am far from being an expert, but I can confirm that there exist number fields $K\neq\mathbb{Q}$ which have no nontrivial unramified extensions. For example, imaginary quadratic number fields of class number one have this property. See this paper by Yamamura for more examples and background information.
On the other hand, I don't understand your main question. What do you mean by "reconstructing" the Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$?
Best Answer
Let me first note that there is a slight ambiguity when one says "ramified only at 2". Strictly speaking, that means that the extension is unramified at every place of $\mathbb Q$ except 2, including infinity. The latter mean that the extension is totally real. Often, however, "ramified only at 2" means "ramified only possibly at 2 and $\infty$", and it is probably what you mean. Here, to remove ambiguity, for $S$ a finite set of places of $\mathbb Q$, I will use "ramified only at $S$" in the strict sense.
That being said, the short answer is that whatever the finite set $S$, there are strong restrictions on the finite groups $G$ that can appear as the Galois group of an extension of $\mathbb Q$ unramified outside at $S$, but that in general, to "describe" all these restrictions (we may mean different things by that) is in general an open problem. To see what kind of restrictions can appear, let us consider several situation, from the very particular to the general.
If $S=\emptyset$ or $S=\{\infty\}$, then Minkowski's theorem tell that there is no nontrivial extension unramified outside $S$, so the only possible Galois group $G$ is the trivial one. A very strong restriction indeed.
If $S$ is any set, but you try to determine what abelian $G$ may appear, then the answer is given by class field theory. Precisely, if $S=\{\ell_1,\dots,\ell_k,\infty\}$, then the abelian group which appear are the ones that are quotient of $\mathbb Z_{ell_1}^\ast \times \dots \mathbb Z_{\ell_k}^\ast$, and it is obvious that many abelian groups are not of this type (e.g. $\mathbb Z/\ell \mathbb Z$ for $\ell > \ell_1,\dots,\ell_k$). If $S=\{\ell_1,\dots,\ell_k\}$, replace $\mathbb Z_{ell_1}^\ast \times \dots \mathbb Z_{\ell_k}^\ast$ by its quotient by the diagonal subgroup $\{1,-1\}$.
If $S$ is any set, and you're interested in groups $G$ that are $p$-groups (for a $p$ that may or may nor be in $S$), then again, class field theory can help you because of Frattini's argument saying that a $p$-group is generated by any set that maps subjectively to the maximal abelian $p$-torsion quotient of $G$. So by the above, the $p$-groups $G$ appearing this way have a system of generators with less elements than the dimension over $\mathbb F_p$ of the maximal $p$-torsion quotient of $\mathbb Z_{ell_1}^\ast \times \dots \mathbb Z_{\ell_k}^\ast$. Is that all? No. We can describe in some cases all the $p$-group $G$ appearing, but not in all. For instance, with $S=\{2,\infty\}$, and $p=2$, the groups $G$ appearing are exactly the $2$-groups having a system of two generators, one of them being of square 1. For $S=\{2\}$, the only $2$-groups appearing are the cyclic groups. In general, the case $p \in S$ is better understood than the case $p \not \in S$. In the latter case, it is a folklore conjecture, on which not much is known, that only finitely many $p$-groups $G$ appearing (one can check readily that the conjecture is true for abelian $p$-groups, by the above paragraph).
If now we consider general finite groups $G$, well, the above shows that there are restrictions, but determining them all is largely open. For instance, a conjecture of Shafarevich states that there is an $n=n(S)$ such that every group $G$ which is the Galois group of an extension unramified outside $S$ has a system of generators with less than $n$ elements. But this is open in every non-trivial case.
Let me also mention a simple but too little known result of Serre, even if it is not strictly speaking part of your question: [Edit: Sorry, I messed up the statement of the result. Here is the right version]. If every finite group is a Galois group over $\mathbb Q$ (if the inverse galois problem has a positive solution) then every finite group is the Galois group of an extension of $\mathbb Q$ unramified at infinity -- that is, a real extension. In other words, according to the inverse Galois problem, there should need no restriction of $G$ in the case where $S$ is the set containing all finite places, but not the place at infinity.