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 $( m_n )$.