Relationship between NP-hard and undecidable problems

Undecidable = unsolvable for some inputs. No matter how much (finite) time you give your algorithm, it will always be wrong on some input.

NP-hard ~= super-polynomial running time (assuming P != NP). That's hand-wavy, but basically NP-hard means it is at least as hard as the hardest problem in NP.

There are certainly problems that are NP-hard which are not undecidable (= are decidable). Any NP-complete problem would be one of them, say SAT.

Are there undecidable problems which are not NP-hard? I don't think so, but it isn't easy to rule it out - I don't see an obvious argument that there must be a reduction from SAT to all possible undecidable problems. There could be some weird undecidable problems which aren't very useful. But the standard undecidable problems (the halting problem, say) are NP-hard.

An NP-hard is a problem that is at least as hard as any NP-complete problem.

Therefore an undecidable problem can be NP-hard. A problem is NP-hard if an oracle for it would make solving NP-complete problems easy (i.e. solvable in polynomial time). We can imagine an undecidable problem such that, given an oracle for it, NP-complete problems would be easy to solve. For example, obviously every oracle that solves the halting problem can also solve an NP-complete problem, so every Turing-complete problem is also NP-hard in the sense that a (fast) oracle for it would make solving NP-complete problems a breeze.

Therefore Turing-complete undecidable problems are a subset of NP-hard problems.