a)

${\left(\frac{1}{16}\right)}^{\frac{1}{4}}=\sqrt[4]{\frac{1}{16}}=\frac{1}{2}$

${\left(\frac{1}{64}\right)}^{\frac{2}{3}}={\left(\frac{\sqrt[3]{1}}{64}\right)}^2={\left(\frac{1}{4}\right)}^2=\frac{1}{16}$

${32}^{\frac{3}{5}}={\left(\sqrt[5]{32}\right)}^3=2^3=8$

$8^{\frac{2}{3}}={\left(\sqrt[3]{8}\right)}^2=2^2=4$

$8>4>\frac{1}{2}>\frac{1}{16}$

${32}^{\frac{3}{5}}>8^{\frac{2}{3}}>{\left(\frac{1}{16}\right)}^{\frac{1}{4}}>{\left(\frac{1}{64}\right)}^{\frac{2}{3}}$