Prerequisites - P.2 - Exponents and Radicals - Exercises - Page 25: 61

$a]$ $\frac{1}{8}$ $b]$ $\frac{27}{8}$

$a]$ $(32)^{\frac{-3}{5}}=$ $(2)^{5 \times \frac{-3}{5}} =$ $(2)^{(-3)}=$ $\frac{1}{8}$ $b]$ $(\frac{16}{81})^\frac{-3}{4}=$ $(\frac{2}{3})^{4\times\frac{-3}{4}}=$$(\frac{2}{3})^{-3}=$ $\frac{27}{8}$