$(\frac{x+2}{2x-4}+\frac{2-x}{4+2x}+\frac{8}{x^2-4}):\frac{4}{x-2}=$

$=(\frac{x+2}{2\left(x-2\right)}+\frac{2-x}{2\left(2+x\right)}+\frac{8}{\left(x+2\right)\left(x-2\right)})·\frac{x-2}{4}=$

$=\left(\frac{\left(x+2\right)\left(x+2\right)+\left(x-2\right)\left(2-x\right)+16}{2\left(x+2\right)\left(x-2\right)}\right)·\frac{x-2}{4}=$

$=\left(\frac{x^2+4x+4-x^2+4x-4+16}{2\left(x+2\right)\left(x-2\right)}\right)·\frac{x-2}{4}=$

$=\left(\frac{8x+16}{2\left(x+2\right)\left(x-2\right)}\right)·\frac{x-2}{4}=$

$=\left(\frac{8\left(x+2\right)}{2\left(x+2\right)\left(x-2\right)}\right)·\frac{x-2}{4}=1$