Describe the groups of homomorphisms of the given abelian groups In algebra we have the following problem to solve:

Describe the groups of homomorphisms of abelian groups.
(a) $\textrm{Hom}(\mathbb{Q} / \mathbb{Z}, \mathbb{Q})$
(b) $\textrm{Hom}(\mathbb{Q}, \mathbb{Q} / \mathbb{Z})$

So first of all, I'm not sure, if I really understand what to do. I assume that I just have to describe the group $\textrm{Hom}(\mathbb{Q/Z,Q})$,
This means that I have to define a group operation and then prove that it satisfies the group axioms, like closedness, associativity, existence of inverses element, existence of an identity element, etc. Is this idea correct?
If so, I would define the operation like this: let $f,g \in \textrm{Hom}(\mathbb{Q/Z,Q})$ then: $$(f\odot g)(x):=f(x)+g(x), \qquad \forall x\in \mathbb{Q}.$$
to check that this operation is closed, we have to show that $$f\odot g \in \textrm{Hom}(\mathbb{Q/Z,Q}), \qquad \forall f,g \in \textrm{Hom}(\mathbb{Q/Z,Q})$$
Proof: Let $f,g \in \forall f,g \in \textrm{Hom}(\mathbb{Q/Z,Q})$. Then,
$$\begin{array}{rclcl}
(f\odot g)(x+y)
    &=& f(x+y)+g(x+y) & \qquad & \text{(by def of $\odot$)} \\
    &=& f(x)+f(y)+g(x)+g(y) & & \\
    &=& f(x)+g(x)+f(y)+g(y) & & \\
    &=& (f\odot g)(x)+(f\odot g)(y). & & \text{($\forall x,y \in \mathbb{Q/Z}$)} 
\end{array}$$
This follows only if $\mathbb{Q}$ is abelian, which is obviously true. Also from this fact follows that: $$(f\odot g)(x)=f(x)+g(x)=g(x)+f(x)=(g\odot f)(x), \qquad \forall x \in \mathbb{Q/Z},$$
and hence $\textrm{Hom}(\mathbb{Q/Z,Q})$ is abelian.
It remains only to show that there exist inverses and an identity element.
As identity element we could take the embedding  $e: \mathbb{Q/Z} \hookrightarrow \mathbb{Q}.$
For the inverse I did this:
$(f\odot g)(x)=e(x)=x$, $\forall x \in \mathbb{Q/Z}$ , so
$$(f\odot g)(x)=f(x)+g(x)=x$$
and so
$f(x)^{-1}:=g(x)=x-f(x)$ should be the inverse.
Is this correct, or am I completely on the wrong trail?
 A: If $A$ and $B$ are abelian groups, then $\operatorname{Hom}(A,B)$ is an abelian group under the obvious operations and I don't think you have to discuss them, because it's general knowledge.
If $A$ is a torsion abelian group and $B$ is a torsion free abelian group, then $\operatorname{Hom}(A,B)=\{0\}$ and this settles the first part. Why is that? If $f\colon A\to B$ is a homomorphism and $x\in A$, choose $n>0$ such that $nx=0$; then $0=f(0)=f(nx)=nf(x)$; since $B$ is torsion free, this implies $f(x)=0$.
A good starting point for the second part is to consider the exact sequence
$$
\def\Z{\mathbb{Z}}\def\Q{\mathbb{Q}}
\def\Hom{\operatorname{Hom}}\def\Ext{\operatorname{Ext}}
0\to\Z\to\Q\to\Q/\Z\to0
$$
and apply the $\Hom(\Q,-)$ functor, which gives the exact sequence
$$
0=\Hom(\Q,\Z)\to\Hom(\Q,\Q)\to\Hom(\Q,\Q/\Z)\to\Ext(\Q,\Z)\to\Ext(\Q,\Q)=0
$$
Now, what's $\Ext(\Q,\Z)$? It's isomorphic to the additive group of the real numbers, as shown by Wiegold (Bull. Austral. Math. Soc. 1 (1969), 341–343).
Since $\Hom(\Q,\Q)\cong\Q$ is divisible, the sequence splits, so we get that
$$
\Hom(\Q,\Q/\Z)\cong \Q\oplus\mathbb{R}\cong\mathbb{R}.
$$
