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a)
$\left(x+2\right)^2=\left(x-4\right)\left(x+4\right)$
$x^2+4x+4=x^2-16$
$4x=-20$
$x=-5$
b)
$\left(x+5\right)^2-\left(x-2\right)^2=0$
$x^2+10x+25-\left(x^2-4x+4\right)=0$
$x^2+10x+2=x^2-4x+4$
$6x=2$
$x=\frac{1}{3}$