$a)\frac{-\sqrt[3]{-8}\cdot \sqrt[5]{-32}+\sqrt{2}\cdot \sqrt{2}}{\sqrt{\frac{16}{3}}\cdot \sqrt[3]{-27}\cdot \sqrt{3}}=$     $\frac{-\left(-2\right)\cdot \left(-2\right)+\sqrt{4}}{\frac{\sqrt{16}}{\sqrt{3}}\cdot \left(-3\right)\cdot \sqrt{3}}=$ $\frac{-\left(-2\right)\cdot \left(-2\right)+\sqrt{4}}{\frac{\sqrt{16}}{\sqrt{3}}\cdot \left(-3\right)\cdot \sqrt{3}}=$    $\frac{-4+2}{\frac{4}{\sout{\sqrt{3}}}\cdot \left(-3\right)\cdot \sout{\sqrt{3}}}=\frac{-2}{-12}=\frac{1}{6}$   

$b)\frac{\sqrt[3]{-27}\cdot \sqrt{\frac{16}{9}}+\sqrt[7]{-1}\cdot \sqrt{25}:\sqrt[3]{-\frac{125}{64}}}{\sqrt[3]{343}=}$    $=\frac{-\sout{3}\cdot \frac{4}{\sout{3}}+\left(-1\right)\cdot 5:\left(-\frac{5}{4}\right)}{7}=$    

$=\frac{-4+\left(-5\right)\cdot \left(-\frac{4}{5}\right)}{7}=$     $\frac{-4+4}{7}=\frac{0}{7}=0$