解：$(1)\begin{matrix} & \dfrac{2}{m−2}− \dfrac{8}{{m}^{2}−4}= \dfrac{2}{m−2}− \dfrac{8}{(m+2)(m−2)}= \dfrac{2(m+2)−8}{(m+2)(m−2)}= \dfrac{2(m−2)}{(m+2)(m−2)}= \dfrac{2}{m+2} \\ & \end{matrix}$；

$(2)( \dfrac{1}{x+1}+ \dfrac{{x}^{2}−2x+1}{{x}^{2}−1})÷ \dfrac{x−1}{x+1}$，其中$x=2$．

$=( \dfrac{1}{x+1}+ \dfrac{x−1}{x+1})⋅ \dfrac{x+1}{x−1}= \dfrac{x}{x+1}− \dfrac{x+1}{x−1}= \dfrac{x}{x−1}$

当$x=2$时原式$=2$．