If P = NP, then how not cannot solve NP-hardness (the one that doesn't intersect with NP-complete) in polynomial-time?
You have your reductions the wrong way around. A problem&nbsp;$X$ is NP-hard if every problem in&nbsp;NP can be reduced to&nbsp;$X$. That is, saying that $X$&nbsp;is NP-hard means that, if we had an efficient algorithm for&nbsp;$X$, we would have an efficient algorithm for every problem in&nbsp;NP: $X$&nbsp;is easy implies all of&nbsp;NP is easy.

So, suppose that P$\,=\,$NP. Then, all of NP is easy, but that tells us nothing about&nbsp;$X$, because the implication goes the wrong way. Maybe $X$&nbsp;is easy; maybe it's hard. In fact, some NP-hard problems are definitely hard: the halting problem is NP-hard but it's undecidable, regardless of what the relationship between P and&nbsp;NP turns out to be.