You might introduce a Grothendieck topology on the Galois subextensions of $L/K$; then the Galois group is indeed a sheaf.
I just want to remark the following: There is a natural homeomorphism between $Gal(L/K)$ and $Spec(L \otimes_K \overline{K})$. Namely, if $\sigma \in Gal(L/K)$, then the kernel of $L \otimes_K \overline{K} \to L \otimes_K \overline{K} \subseteq \overline{K} \otimes_K \overline{K} \to \overline{K}$ is a prime ideal of $L \otimes_K \overline{K}$. This is easily checked to be a homeomorphism if $L/K$ is finite, and then also in general. In particular, you can endow the profinite space $Gal(L/K)$ with a sheaf so that we get an affine scheme. If $L'/K$ is a Galois subextension of $L/K$, then the restriction $Gal(L/K) \to Gal(L'/K)$ comes from the inclusion $L \otimes_K \overline{K} \subseteq L' \otimes_K \overline{K}$. This provides another reason why $Gal(-/K)$ is a "sheaf", namely because $Spec$ commutes with filtered direct limits.
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.
Best Answer
Any profinite simple group is finite, since it has nontrivial finite quotients (the conjugates of a finite index subgroup intersect in a finite index subgroup).