These are the (torsion) cotorsion groups. The following follows from a theorem of Baer:
A torsion abelian group is cotorsion if and only it is direct sum of a divisible torsion abelian group and an abelian group of bounded exponent.
The original paper of R. Baer is "The subgroup of the elements of finite order of an abelian group", Ann. of Math. 37 (1936), 766-781. (See in particular Theorem 8.1.)
[I have made Baer's paper available here. --PLC]
See:
V. S. Guba, Finitely generated complete groups, Izv. Akad. Nauk SSSR Ser. Mat. 50 (1986),
883-924.
for an interesting 2-generated example. (Furthermore, roots in Guba's examples are unique.) Note that a group $G$ is called complete if for any non-trivial word $u(x_1,\cdots,x_m)$ and every $g\in G$, the equation $u(x_1,\cdots,x_m)=g$ has a solution. In your case, the words $u$ are of the form $x^n$. I do not know if Guba's group is simple, but it does have a (nontrivial) simple quotient, which is necessarily verbal, and, in particular, divisible.
None of these groups is finitely-presented and, I think, the problem of existence of fp divisible groups is open. The philosophical reason why is the following. Call an infinite finitely-generated group "exotic" if it satisfies some bizarre, seemingly impossible property, e.g., being a torsion group, containing very few conjugacy classes, being divisible, etc. The most common method for constructing exotic groups $G$ is by direct limit of a sequence of (relatively) hyperbolic groups $G_k$ which are quotients of a single group $G_0$. If $G$ were finitely presented, it would be isomorphic to one of the groups $G_k$ and, hence, non-exotic.
Mark Sapir will probably have more comments on this.
Best Answer
By "uniquely divisible", I guess you mean torsion-free and divisible, i.e., $Q$ is a $\mathbb{Q}$-vector space. If so, the answer to your question is that every such exact sequence splits.
It is enough to show $\mathrm{Ext}(Q, \mathbb{Z}/n) = 0$. We have an exact sequence
$$\hom(Q, \mathbb{Z}/n) \to \mathrm{Ext}(Q, \mathbb{Z}) \stackrel{ \cdot n}{\to} \mathrm{Ext}(Q, \mathbb{Z}) \to \mathrm{Ext}(Q, \mathbb{Z}/n) \to \mathrm{Ext}^2(Q, \mathbb{Z})$$
where the first and last terms are zero. From Hilton-Stammbach, A Course in Homological Algebra exercise 6.1 (p. 109), we have that $\mathrm{Ext}(Q, \mathbb{Z})$ is divisible if $Q$ is torsionfree, and $\mathrm{Ext}(Q, \mathbb{Z})$ is torsionfree if $Q$ is divisible. Hence in our case $\mathrm{Ext}(Q, \mathbb{Z})$ is torsionfree and divisible, hence a $\mathbb{Q}$-vector space, and so multiplication by $n$ is an isomorphism and the claim follows.