Fundamental group of 3-manifold with boundary Is it true that any finitely presented group can be realized as fundamental group of compact 3-manifold with boundary?
 A: If a finite group $G$ is (isomorphic to) the fundamental group of a three-manifold, then $G$ embeds in $\mathrm{SO}(4)$.
To see this, suppose that $M$ is a compact, connected three-manifold, possibly with boundary, having $G = \pi_1(M)$ finite. If $M$ has two-sphere boundary components, we can cap them off without changing the fundamental group.  Since free products of non-trivial groups are always infinite, we deduce that $G$ is freely indecomposible.  Appealing to the Poincaré conjecture [solved by Perelman], we have that $M$ is irreducible: all two-spheres in $M$ bound three-balls.
If $M$ is non-orientable, then a theorem of Livesay implies that $M$ is homeomorphic to the real projective plane crossed with a unit interval.
Thus we reduce to the case where $M$ is compact, oriented, connected, irreducible, and has finite fundamental group.  All boundary components of $M$ are now oriented, and of genus at least one.  Applying "one-half lives, one-half dies" we find that $M$ has no boundary components.  Appealing to the elliptic part of the geometrisation conjecture [solved by Perelman] we find that the universal cover of $M$ is the three-sphere, and the deck group, and thus $G$, is (conjugate to a) subgroup of $\mathrm{SO}(4)$. QED
A quick google search finds a paper of Zimmermann giving a readable introduction to the finite subgroups of $\mathrm{SO}(4)$ - see section three of that paper.  One then has to determine which of these act freely.  Finally, there is another approach to this problem via spherical Seifert fibered spaces.
As a concrete example of a finite group that is not a three-manifold group, consider $\mathbb{Z}_p \times \mathbb{Z}_q$ where $p$ and $q$ are not coprime.  This is a isomorphic to a subgroup of $\mathrm{SO}(4)$, but it cannot act freely.  [See Theorem 9.14 in Hempel's book for a proof that, in this special case, avoids geometrisation.]
A: No.  The Baumslag solitar groups $\langle a, b | ab^m a^{-1} = b^n \rangle$ are not $3$-manifold groups when $m \neq n$.  
See
Heil, Wolfgang H. Some finitely presented non-$3$-manifold groups. Proc. Amer. Math. Soc. 53 (1975), no. 2, 497--500. 
(See also Peter Shalen, Three-Manifolds and Baumslag-Solitar groups.
Topology Appl. 110 (2001), 113--118) 
A: A couple of extra points.
Any compact 3-manifold with boundary $M$ can be doubled to give a closed 3-manifold $D$.  As $M$ is a retract of $D$, it follows that $\pi_1(M)$ injects into $\pi_1(D)$.  Therefore, any "poison subgroup" (such as the Baumslag--Solitar groups that Autumn mentions above) applies just as well to compact 3-manifolds as closed 3-manifolds.
Other classes of poison subgroups can be constructed from cohomological conditions.  The Kneser--Milnor Theorem implies that any closed, irreducible 3-manifold with infinite fundamental group is aspherical.  It follows that any freely indecomposable infinite group with cohomologial dimension greater than 3 cannot be a subgroup of a closed 3-manifold (and hence of a compact 3-manifold, by the previous paragraph).
EDIT:
Oh, and yet another source of poison subgroups comes from Scott's theorem that 3-manifold groups are coherent, meaning that every finitely generated subgroup is finitely presented.  This rules out subgroups like $F\times F$ (where $F$ is a free group), which is not coherent.
A: I recently heard of a result due to Aitchison and Reeves which shows that any finitely presented group arises as the fundamental group of a 3-dimensional orbifold (where fundamental group means the topological and not the orbifold fundamental group). In fact, they say that the orbifold can be taken to be the quotient of a closed oriented hyperbolic 3-manifold by an isometric involution with isolated fixed points, all modelled on $x\mapsto -x$.
(I'm certainly no expert on this topic, just passing on what I heard.) 
