解：$(1)$当$m=2$，$n=1$时，$(m+n)^2=(2+1)^2=9$
$\mathrm {m^2}+2mn+n^2=2^2+2×2×1+1^2=9$
$(2)(m+n)^2=\mathrm {m^2}+2mn+n^2$
$(3)$当$m=4$，$n=-2$时，$(m+n)^2=[4+(-2)]^2=2^2=4$
$\mathrm {m^2}+2mn+n^2=4^2+2×4×(-2)+(-2)^2=16+(-16)+4=4$
∴$(m+n)^2=\mathrm {m^2}+2mn+n^2$仍成立
$(4)$根据$(2)$的结论得$\mathrm {m^2}+2mn+n^2=(m+n)²$
$ $当$m=0.125$，$n=0.875$时
$\mathrm {m^2}+2mn+n^2=(m+n)^2=(0.125+0.875)^2=1$
