Homomorphisms and Isomorphisms · Challenge problem

Yes, via the isomorphism $x \mapsto 2^x$.

Yes, since both are countable.

No, because $\mathbb Q^{+}$ is not abelian.

No: every element of $(\mathbb Q,+)$ is $y+y$ for some $y$, but $2$ is not a perfect square in $\mathbb Q^{+}$.