解:$(1)$原式$=(x^{2m})^3+(y^n)^6-x^{2m}·y^n·x^{4m}·y^{5n}$
$ =x^{6m}+y^{6n}-x^{6m}·y^{6n}$
$ =(x^{3m})^2+(y^{3n})^2-(x^{3m}·y^{3n})^2$
$ =4^2+5^2-(4×5)^2$
$ =16+25-400=-359$
$ (2)$因为$1+2+3+\dots +n=m,$
$ $所以$(ab^n)·(a^2b^{n-1})·\dots ·(a^{n-1}b^2)·(a^nb)$
$ =a^{1+2+\dots +n}·b^{n+(n-1)+\dots +1}$
$ =a^mb^m$