Benjamin's answer can be slightly strengthened, in the following way:

**Proposition:** There is a fixed finitely presented group $H$, and fixed finite set of words $S$ on the generators of $H$ for which $\langle S\rangle \leq H$ has solvable subgroup membership problem, such that the problem of determining if the subgroup generated by one word $\langle w \rangle$ intersects $\langle S \rangle$ non-trivially in $H$ is algorithmically undecidable.


**Proof:** There exists a finitely presented group $G=\langle X|R \rangle$ for which the word problem is solvable, but for which the *torsion problem*, of determining if a word has infinite order, is unsolvable. This is a consequence of Theorem A in D. Collins, *The word, power, and order problem in finitely presented groups*, in *Word Problems*, eds. W. W. Boone, F. B. Cannonito and R. C. Lyndon, 401–420 (1973). (On top of this theorem, one needs the additional observation that having solvable word problem means one can effectively compute the order of a torsion element). So now take the group $H$ to be $F_{X} \times F_{X}$, and take the set $S$ to be the finite set $\{(x_{1}, x_{1}), \ldots, (x_{n}, x_{n}), (r_{1}, 1) \ldots, (r_{m}, 1)\}$, and thus $\langle S \rangle =\{(u,v) \in F_{X}\times F_{X}\ |\ u=v \ \ in \ \ G \}$. Since $G$ has solvable word problem, we have that $\langle S \rangle \leq H$ has solvable membership problem. Moreover, as Benjamin observed, for an arbitrary word $z \in G$, we have $\langle (1,z) \rangle$ intersects $\langle S \rangle$ trivially if and only if $z$ has infinite order in $G$ (the latter being undecidable).

I suppose the above should really be a comment to Benjamin's answer, but I can't add comments yet as I have <50 reputational points. My apologies.

-Maurice