Decidability of knot equivalence in general 3-manifolds? Surface equivalence? Given a closed orientable 3-manifold $M^3$ and two knots $K_1$ and $K_2$ in $M$, is there an algorithm to decide if $K_1$ and $K_2$ are isotopic?  Is there an algorithm to decide if there is a homeomorphism $f: M \to M$ with $f(K_1) = K_2$?
I would expect these questions to be involved (at least as difficult as the recognition problem for 3-manifolds) but I thought that the answer is possibly a folklore consequence of geometrization.  The special case of the first question for $K_1$ unknotted follows from a result of Hass and Lagarias (see Theorem 1.2).  The case where $M = S^3$ is a result of Waldhausen using normal surface theory.
As an aside, I don't know anything about the corresponding questions where instead of considering knots we consider closed embedded surfaces $F_1$ and $F_2$.  Even in the case where $M = S^3$ (so only the isotopy question is relevant) the genus of $F_1$ and $F_2$ is greater than one, this is a total mystery to me.  By looking at tori bounding regular neighborhoods of knots, this is strictly more difficult than the corresponding questions for knots.  Are there results in this direction?
 A: Regarding the second question.  This reduces to the homeomorphism problem for three manifolds, which is known due to geometrisation.  Here is a sketch.  Fix $M$ as well as the knots $K$ and $L$.  Let $X_K$ and $X_L$ be the knot complements.  We mark the boundary of each with the meridional slope of $K$ and $L$, respectively. We ask our solution to the homeomorphism problem if $X_K$ and $X_L$ are homeomorphic.  If the answer is "no" then we are done.  If the answer is "yes" we ask for all homeomorphisms from $X_K$ to $X_L$.  For each of these we restrict to the boundary and see if the resulting map takes the meridian of $K$ to that of $L$. If any do, the answer is "yes".  If none do, the answer is "no".
[Edit - as pointed out in the comments, there may be infinitely many homoeomorphisms between $X_K$ and $X_L$.  This happens, for example, when $K$ and $L$ are both unknots in the three-sphere.  In any case, the self-homeomorphisms of $X_K$ restrict to give a virtually cyclic group of homeomorphisms of $\partial X_K$.  When this group is infinite, the cyclic subgroup is generated by a Dehn twist about the slope that dies in homology.]
Regarding the first question.  If $M = S^3$ then again this reduces to the homeomorphism problem for three-manifolds, by appealing to Property P (due to Gordon-Luecke).  Some fiddling about is needed to deal with mirror images.
I don't know the status of the first question for other manifolds off the top of my head.  I'll report back if anything occurs to me. :)
