The question is about characterising the sets $S(K)$ of primes which split completely in a given galoisian extension $K|\mathbb{Q}$.  Do recent results such as Serre's modularity conjecture (as proved by Khare-Wintenberger), or certain cases of the Fontaine-Mazur conjecture (as proved by Kisin), have anything to say about such subsets, beyond what Class Field Theory has to say ?

I'll now introduce some terminology and recall some background.

Let $\mathbb{P}$ be the set of prime numbers.  For every galoisian extension $K|\mathbb{Q}$, we have the subset $S(K)\subset\mathbb{P}$ consisting of those primes which split (completely) in $K$.  The question is about characterising such subsets; we call them galoisian subsets.

If $T\subset\mathbb{P}$ is galoisian, there is a unique galoisian extension $K|\mathbb{Q}$ such that $T=S(K)$, cf. Neukirch (13.10).  We say that $T$ is abelian if $K|\mathbb{Q}$ is abelian.

As discussed here  recently, a subset $T\subset\mathbb{P}$ is abelian if and only if it is defined by congruences.  For example, the set of primes $\equiv1\pmod{l}$  is the same as $S(\mathbb{Q}(\zeta_l))$.  "Being defined by congruences" can be made precise, and counts as a characterisation of abelian subsets of $\mathbb{P}$.

Neukirch says that Langlands' Philosophy provides a characterisation of all galoisian subsets of $\mathbb{P}$.  Can this remark now be illustrated by some striking example ?

Addendum (28/02/2010) Berger's recent Bourbaki exposé 1017 arXiv:1002.4111 says that cases of the Fontaine-Mazur conjecture have been proved by Matthew Emerton as well.  I didn't know this at the time of asking the question, and the unique answerer did not let on that he'd had something to do with Fontaine-Mazur...

I think it is easiest to illustrate the role of the Langlands program (i.e. non-abelian class field theory) in answering this question by giving an example.
E.g. consider the Hilbert class field $K$ of $F := {\mathbb Q}(\sqrt{-23})$; this is a degree 3 abelian extension of $F$, and an $S_3$ extension of $\mathbb Q$.  (It is the splitting field of the polynomial $x^3 - x - 1$.)
The 2-dimensional representation of $S_3$ thus gives a representation
$\rho:Gal(K/{\mathbb Q}) \hookrightarrow GL_2({\mathbb Q}).$
A prime $p$ splits in $K$ if and only if $Frob_p$ is the trivial conjugacy class
in $Gal(K{\mathbb Q})$, if and only if $\rho(Frob_p)$ is the identity matrix, if and only
if trace $\rho(Frob_p) = 2$. (EDIT: While $Frob_p$ is a 2-cycle, resp. 3-cycle, if and only if $\rho(Frob_p)$ has trace 0, resp. -1.)
Now we have the following reciprocity law for $\rho$: there is a modular form $f(q)$, in fact
a Hecke eigenform, of weight 1 and level 23, whose $p$th Hecke eigenvalue gives
the trace of $\rho(Frob_p)$.  (This is due to Hecke; the reason that Hecke could handle
this case is that $\rho$ embeds $Gal(K/{\mathbb Q})$ as a dihedral
subgroup of $GL_2$, and so $\rho$ is in fact induced from an abelian character of the
index two subgroup $Gal(K/F)$.)
In this particular case, we have the following explicit formula:
$$f(q) = q \prod_{n=1}^{\infty}(1-q^n)(1-q^{23 n}).$$
If we expand out this product as $f(q) = \sum_{n = 1}^{\infty}a_n q^n,$
then we find that $trace \rho(Frob_p) = a_p$ (for $p \neq 23$),
and in particular, $p$ splits completely in $K$ if and only if $a_p = 2$.
(For example, you can check this way that the smallest split prime is $p = 59$;
this is related to the fact that $59 = 6^2 + 23 \cdot 1^2$.).
(EDIT: While $Frob_p$ has order $2$, resp. 3, if and only if $a_p =0$, resp. $-1$.)
So we obtain a description of the set of primes that split in $K$ in terms of
the modular form $f(q)$, or more precisely its Hecke eigenvalues (or what amounts
to the same thing, its $q$-expansion).
The Langlands program asserts that an analogous statement is true for any
Galois extension of number fields $E/F$ when one is given a continuous
representation $Gal(E/F) \hookrightarrow GL\_n(\mathbb C).$  This is known
when $n = 2$ and either the image of $\rho$ is solvable (Langlands--Tunnell) or $F = \mathbb Q$ and $\rho(\text{complex conjugation})$ is non-scalar (Khare--Wintenberger--Kisin).
In most other contexts it remains open.
