Actually, I am harder-pressed to say anything the supposedly simpler question. For Question 1, I would just add that we need the existence of an infinite computably enumerable set of primes satisfying FLT, and that this ought to suffice. (Since the set P of primes NOT satisfying FLT is obviously $\exists$-definable and hence c.e., another way to say this is that we need to show that P is not a simple set.) I would not have guessed that any odd prime was known to satisfy FLT, up until Wiles's proof, but I'm ready to be corrected on that.
For Question 2, I do think that one could come up with a proof avoiding FLT entirely. I can't do so myself; I'm in computability, and when I was thinking about computable categoricity for fields of infinite transcendence, I realized that we needed multivariable polynomials with known finite numbers of rational solutions, and I thought of the Fermat polynomials because I didn't know any other candidates. (Of course, this condition was not all that was needed, but clearly it's necessary.) I'm still a computability theorist, and I still don't know any other candidates, but field theorists have told me that they could come up with other such polynomials fairly readily. Whether those would satisfy the more difficult requirements (basically Theorem 3.1 in the paper) is not so clear, but I suspect that it can be done with other polynomials. Bjorn Poonen suggested at one point that the Fermat polynomials were actually a bad choice, because their symmetry creates an extra solution whenever one adds a single transcendental solution to the field.
As a related question: is there a computably stable field of infinite transcendence degree? A computable structure $\mathcal{A}$ is computably stable if, for every computable structure $\mathcal{B}$, every isomorphism from $\mathcal{A}$ onto $\mathcal{B}$ is computable. (Example: $\mathbb{Z}$ under the successor function.) A common way to build computably stable structures is to make them computably categorical and rigid, i.e. with no nontrivial automorphisms, so that the isomorphism from $\mathcal{A}$ onto any computable copy must be unique (by rigidity), hence computable (by categoricity). I would conjecture that it is possible to mimic the construction in the Fermat paper, with different polynomials, and to get a computably stable field of infinite transcendence degree, but I certainly don't know offhand what polynomials one might use.
Actually Ken Ribet proved that if $K$ is a number field and $K(\mu_{\infty})$ is its infinite cyclotomic extension generated by all roots of unity then for every abelian variety $A$ over $K$ the torsion subgroup of $A(K(\mu_{\infty}))$ is finite:
http://math.berkeley.edu/~ribet/Articles/kl.pdf .
On the other hand, Alosha Parshin conjectured that if $K_{p}$ is the extension of $K$ generated by all $p$-power roots of unity (for a given prime $p$) then the set $C(K_{p})$ is finite for every $K$-curve $C$ of genus $>1$:
http://arxiv.org/abs/0912.4325 (see also http://arxiv.org/abs/1001.3424 ).
Best Answer
This is not quite an answer, but not quite a comment either.
We can at least show that $X(({\mathbb Q}^{\text{ab}})^{\text{ab}})$ is infinite (where $X$ is the quintic Fermat curve). There are in fact (at least) two ways of showing this.
The second construction obviously generalizes to arbitrary Fermat curves. In my originial post, I claimed that the first one does, too. But this seems to be wrong: the quotient of the degree $n$ Fermat curve $x^n + y^n + z^n = 0$ by the obvious $S_3$-action has positive genus as soon as $n \ge 6$, so the quotient by the cyclic group is no longer obviously hyperelliptic. (It is true, however, that this curve maps to the hyperelliptic curve $y^2 = x^n + \frac{1}{4}$, but this is via the quotient w.r.t. the action of $\mu_n$ such that $\zeta$ sends $(x:y:z)$ to $(\zeta x : \zeta^{-1} y : z)$. This would lead to a weaker conclusion, replacing the double by the triple abelian closure of $\mathbb Q$. I had mixed up the two quotients.)
Note that these constructions use the fact that $X$ has (many) nontrivial automorphisms. So perhaps another interesting question is the following.
Let $C$ be a nice (smooth, projective, geometrically irreducible) curve over $\mathbb Q$. Can we at least show that $C({\mathbb Q}^{\text{sol}})$ is infinite?
Here ${\mathbb Q}^{\text{sol}}$ denotes the union of all finite Galois extensions of $\mathbb Q$ with solvable Galois group.
(Originally the question was formulated for (sufficiently generic) curves of genus 3. However, as René pointed out, we find lots of quartic points by intersecting the plane quartic $C$ with rational lines, and all quartic fields are contained in ${\mathbb Q}^{\text{sol}}$.)
EDIT: The question whether solvable points exist has been studied, see for example a recent preprint by Trevor Wooley and the references given there, in particular this paper by Ambrus Pál.