2) A problem is NP complete if it can be proved to have been solved.
This is not correct. A problem is NP complete if it is both NP-hard (at least as _hard_ as any problem in NP) and it is also in NP (no harder than any problem in NP). That is, it his "hard", but not "too hard." A problem is in NP if there is a polynomial time algorithm that verifies whether a _solution instance_ is correct. For example, the TSP as posed for NP completeness is a decision problem: Yes or no, is there a path of length L or smaller? It is easy to verify if a solution path has length L or smaller in polynomial time, so the TSP decision problem is in NP.
3) It just means there is a polynomial algorithm for that particular problem.
Sorry. Also not true. Again, NP hard means "at least has hard as any problem in NP." So if you find a polynomial time solution to an NP-hard problem, you have in fact collapsed NP into P.
No one so far has also mentioned that NP-hard includes other-than-decision problems. For example, consider General TSP: "return the _shortest_ circuit among cities." This can't be NP-complete because only decision problems with yes/no answers lie in NP. Nonetheless, a polynomal time algorithm for General TSP easily gives you on for the NP-complete yes/no version of TSP as well, so General TSP is by definition NP-hard.