a)
\frac{2\sqrt2 \cdot 16^{-\frac{1}{4}}}{(8^{-\frac{1}{3}} \cdot 4)^3}=\frac{2^1\cdot 2^{\frac{1}{2}}\cdot (2^4)^{-\frac{1}{4}}}{[(2^3)^{-\frac{1}{3}}\cdot 2^2]^3}=
\frac{2^{\frac{3}{2}-1}}{(2^{-1+6})^3}\\=\frac{2^{-\frac{1}{2}}}{2^{15}}=2^{-\frac{1}{2}-15}=2^{-15,5}
b) \frac{2^{\frac{2}{3}}\cdot \sqrt8}{16^{\frac{2}{3}}\cdot \sqrt{32}}= \frac{2^\frac{2}{3}\cdot (2^3)^{\frac{1}{2}}}{(2^4)^{\frac{2}{3}}\cdot (2^5)^\frac{1}{2}}=\frac{2^{\frac{2}{3}+\frac{3}{2}}}{2^{\frac{8}{3}+\frac{5}{2}}}=\frac{2^{\frac{4+9}{6}}}{2^{\frac{16+15}{6}}}=2^{\frac{13}{6}-\frac{31}{6}}=2^{-\frac{18}{6}}=2^{-3}