No. The Rips Construction gives, for any fp group Q, a fp (in fact, C'(1/6)) group G and a surjection G-> Q. The kernel is 2-generated, but it follows from a theorem of Bieri that it is only fp if Q is finite.

References to be added when I have time.

No. Ollivier & Wise's version of the Rips Construction gives, for any finitely presented group $Q$, a finitely presented group $G$ of cohomological dimension 2 and a surjection $G\to Q$ such that the kernel $K$ satisfies:

1. $K$ is finitely generated; and
2. $K$ has Kazhdan's property T, in particular $K$ has at most one end.

Now it follows from Theorem 5.3 of a paper of Bieri that $K$ is only finitely presented if $Q$ is finite.

Note: In my original answer, I only mentioned the unadulterated Rips Construction. Using Ollivier and Wise's version is overkill, but it makes the application of Bieri's theorem cleaner.

I should also mention another, famous and beautiful (though I suppose less general) counterexample. In its simplest cases this example is more elementary.

Given a flag complex $L$, Bestvina & Brady consider the corresponding right-angled Artin group $A_L$ and the kernel $K_L$ of the map $A_L\to\mathbb{Z}$ that sends each generator to $1$. They prove:

1. $K_L$ is finitely generated if and only if $L$ is connected; and
2. $K_L$ is finitely presented if and only if $L$ is simply connected.

So just take $L$ to be your favourite connected, non-simply connected flag complex to construct a counterexample. The square graph with four vertices and four edges is a good choice for $L$, in which case $A_L$ is just the direct product of two copies of the free group on two generators. In this simple case, it's easy to see that $K_L$ is finitely generated; one should be able to prove (though I haven't tried) that $K_L$ is not finitely presented by messing around with some spectral sequences...