Does an undecidable decision problem have a ZFC-independent instance? Is it true that every undecidable decision problem has an instance whose solution is independent of ZFC?
For example, let $G$ be a finitely-presented group with undecidable word problem.  Does there necessarily exist a finite word $w$ over the generators of $G$ such that the statement "$w$ represents the identity" is independent of ZFC?
 A: I think the answer is yes if you believe that ZFC is true.  If all instances of the decision problem were provable or disprovable in ZFC, then you can decide the decision problem for a given instance just by simultaneously searching for a proof and a disproof of the instance in ZFC.  (You need the hypothesis that ZFC is true to show that this actually does decide the problem.)
Edit: Instead of "ZFC is true", maybe a better way to phrase it is this:  Suppose the decision problem is to decide whether or not $\phi(n)$ for $n\in\mathbb{N}$.  Then the hypothesis that all instances are decidable by ZFC is: $\forall n\, \left[(\mathrm{ZFC}\vdash\phi(n))\vee(\mathrm{ZFC}\vdash\neg\phi(n))\right]$.
In order to know that the above procedure actually does decide the problem, you have to know that $\forall n\,\left[ (\mathrm{ZFC}\vdash\phi(n))\rightarrow \phi(n)\right]$ and $\forall n\, \left[(\mathrm{ZFC}\vdash\neg\phi(n))\rightarrow \neg\phi(n)\right]$.  One way to guarantee this would be if you believed that for all sentences $\psi$, $(\mathrm{ZFC}\vdash\psi)\rightarrow\psi$ (this is what I informally meant by "ZFC is true").  Note that this is stronger than "ZFC is consistent", since "ZFC is consistent" is obtained by taking $\psi$ to be False.
