No. First, either a solution has to be found by a nondeterministic TM or a solution and a proof that it is indeed a solution have to be verified by a deterministic TM. (Both variants have to be done in polynomial time.) And then such problem is called to be in NP, not NP-complete.np complete problems are those problems whose solution can be verified in polynomial time by a non deterministic turing machine.
Usually one deals with decidable problems when talking about NP-complete, NP-hard, etc. Thus, solutions to such problems can not only be verified, but also found.np hard problems are the hardest problems whose solution might not be verifiable.
Yes. This can be taken as a definition.also a problem is np hard if there exist np complete problem that is reducible to np harp problem
Basically, a problem P is called NP-hard if every problem in NP is polynomial-time reducible to P. A problem is called NP-complete if it is NP-hard and it itself is in NP. Thus, NP-hard problems can be harder than NP-complete ones.
The Wikipedia article describes this pretty well.