Why are NP problems called that way (and NP-hard and NP-complete)?

Question

Really.. I\'m having the last test for graduation this Tuesday, and that\'s one of the things I just never could understand. I realize that a solution for NP problem can be

Answer 1

But what does determinism has to do with that?

From Wikipedia:

NP is the set of all decision problems for which the 'yes'-answers have efficiently verifiable proofs of the fact that the answer is indeed 'yes'. More precisely, these proofs have to be verifiable in polynomial time by a deterministic Turing machine

Not sure about the history of the names though. EDIT: I have guesses though. What follows is speculation but I don't think it's unreasonable.

NP-Hard is any problem which is at least as hard as the hardest problems in NP. Therefore, it could be said that the problem in question would have NP hardness., hence NP-Hard.

If any single problem in NP-complete can be solved quickly, then every problem in NP can also be solved quickly. Therefore, the complete set of NP problems could be solved.

EDIT2: Wikipedia's Complete (Complexity) article indicates:

a computational problem is complete for a complexity class if it is, in a formal sense, one of the "hardest" or "most expressive" problems in the complexity class