$(\frac{1}{2})^3 = \frac{1}{8}$
$(\frac{1}{2})^0 = 1$
$(\frac{1}{2})^{-1} = 2$
$(\frac{1}{2})^{-2} = 2^2 = 4$
$4 > 2 > 1 > \frac{1}{8}$
Ответ:
$(\frac{1}{2})^{-2} > (\frac{1}{2})^{-1} > (\frac{1}{2})^0 > (\frac{1}{2})^3$

$4^{-1} = \frac{1}{4}$
$4^3 = 64$
$4^0 = 1$
$4^{-2} = (\frac{1}{4})^2 = \frac{1}{16}$
$64 > 1 > \frac{1}{4} > \frac{1}{16}$
Ответ:
$4^3 > 4^0 > 4^{-1} > 4^{-2}$