Is there a manifold with fundamental group $\mathbb{Q}$? It is known that the fundamental group of a locally path connected, path connected compact metric space is finitely presented or uncountable. Furthermore the fundamental group of every manifold is countable so the fundamental group of every compact manifold must be finitely presented.

Q1: Is there a non-compact manifold whose fundamental group is isomorphic to $\mathbb{Q}$, the group of rational numbers?

 

Q2: Or more general, for given countable group $G$, is there a manifold whose fundamental group is isomorphic to $G$?

 A: This was answered in the comments.  The answer to both questions is "Yes".  As Fernando Muro mentioned, any countable group is the fundamental group of a manifold that can be embedded in $\mathbb{R}^5$.
There is an open subset of $\mathbb{R}^3$ with fundamental group $\mathbb{Q}$, the complement of a particular solenoid. 
Let $C_1$ be the complement of an unknotted solid torus. $\pi_1(C_1) = \langle x_1 \rangle$. Consider an unknotted solid torus inside the first which wraps around it twice. Let the complement of this be $C_2$. There is an inclusion $\pi_1(C_1) \hookrightarrow \pi_1(C_2) = \langle x_2 \rangle$ so that $x_1 = x_2^2$, or in additive notation, $x_1 = 2x_2$. 

Let the $n$th solid torus wrap around the inside of the $n-1$st solid torus $m_n=n$ times. Let $C_n$ be the complement of this solid torus. There is an inclusion $\pi_1(C_{n-1}) \hookrightarrow \pi_1(C_n) = \langle x_n \rangle$ so that $x_{n-1} = n x_n$.
The solenoid is the intersection of this infinite sequence of nested solid tori. The intersection of the solenoid with a meridianal disk of the first solid torus is a Cantor set. The solenoid can be viewed as a mapping cylinder of an automorphism of a Cantor set.  The complement $C$ is the union of the complements. Any loop in the complement is contained in some $C_n$. The fundamental group is the direct limit, isomorphic to $\mathbb{Q}$ with $x_n = 1/n!$. 
If you choose $m_n=2$ instead of $m_n=n$, the solenoid complement has a fundamental group isomorphic to the dyadic rationals. You can select other subgroups of $\mathbb{Q}$ by varying the sequence $( m_n )$.
