Does there exist an abelian group with insoluable word problem? Does there exist an abelian group with recursively enumerable presentation and insoluble word problem?
My gut says "of course not!". However, my mind keeps saying "but...doesn't $\mathbb{R}$ have insoluble word problem? Wasn't that what Turing's original paper on computability basically did? And this makes sense, because the reals are uncountable so do not have a recursively enumerable presentation." (I should say that my mind is unwilling to commit to these claims.)
Basically, I have no idea. "Clearly" there exists no such group, but I know that whenever the word "clearly" is used there is something that someone is trying to hide...
 A: In the finitely generated case the answer is NO as pointed out by @Timbuc in the comments. In the infinitely generated case I believe that the answer is YES by the following construction.
Let $K \subseteq \mathbb{N}$ be a recursively enumerable non-recursive set and consider a countable abelian group given by the following presentation (commutators omitted):
$$
 \langle x_1, x_2,\dots \| x_k^k = 1 \mbox{ whenever } k \in K\rangle.
$$
If we could solve word problem in this group then we could decide the set $K$ which would be a contradiction.
A: The question is unclear, since as Derek Holt mentioned, "solvable word problem" has not been defined. If you have a presentation with a sequence of generators, you can ask two questions:
1) whether this group endowed with this generating family has a solvable word problem
2) whether this group is isomorphic to a computable group (i.e. is finite or isomorphic to $\mathbf{N}$ endowed with a computable law; this also means that it admits a generating family (maybe not the initial one) for which the word problem is solvable.
For instance, as a variation of Ferov's answer: let $K\subset\mathbf{N}$ be a recursively enumerable, non-recursive set.
a) Consider the group $\langle x_1,x_2,\dots\mid x_k=1 \forall k\in K\rangle$
then the word problem is not solvable with respect to this generating family. But of course this group is free abelian of countable rank, hence has solvable word problem with respect to a better choice of generating family!
b) If instead we consider, the group $$\langle x_1,x_2,\dots\mid x_j^j=1 \forall j\in\mathbf{N},\;x_k=1\forall k\in K\rangle,$$ then it is isomorphic to $A_K=\bigoplus_{k\in\mathbf{N}-K}\mathbf{Z}/k\mathbf{Z}$. Now assume that $K$ contains all non-primes: then if $A_K$ were computable, we could enumerate all the orders of torsion elements of $A_K$, which is exactly $\mathbf{N}-K$, a contradiction. 
