Is there a statement whose undecidability is undecidable? We know there are statements that are undecidable/independent of ZFC. Can there be a statement S, such that (ZFC $\not\vdash$ S and ZFC $\not\vdash$ ~S) is undecidable?
At least one version of this question has a nearly-trivial answer; if you want to know if there's some statement G such that ZFC (or your PA-compatible logic system of choice) doesn't prove G, ZFC doesn't prove ~G, and we can't prove that ZFC doesn't prove G or not G, then any undecideable statement works!

Specifically, it's impossible for ZFC to prove that it can't prove something, because such a statement is tantamount to the consistency of ZFC; if ZFC were inconsistent then it could prove everything, so proving that there's something that can't be proved is equivalent to proving the consistency of ZFC, and of course this is forbidden by the second incompleteness theorem.
Yes. This is true for any first order statement by Church's theorem, not just statements with respect to ZFC. First order logic is semidecidable, in that every first order statement that can be proven has a sound, complete proof calculus; Valid statements will always be decidable, and can always be proven, though this may take quite a bit of computation.

Satisfiability is undecidable in general however. If a statement is satisfiable, sometimes it can be shown to be satisfiable with finite models, or forcing. If the statement itself and it's negation is demonstrably satisfiable, then you've just proved the statement isn't a universally valid statement. However, one of the result's of Church's theorem proves that there's no algorithm that you can rely on to terminate trying to check for satisfiability of an arbitrary first order statement, because otherwise it would solve the Entscheidungsproblem.