$\text{a)}\frac{4^3\cdot {16}^{\frac{1}{4}}:\sqrt[5]{32}}{{64}^{-\frac{3}{4}}\cdot 8^{\frac{5}{3}}}=\frac{64\cdot {\left(2^4\right)}^{\frac{1}{4}}:2}{{\left(\frac{1}{64}\right)}^{\frac{3}{4}}\cdot {\left(\sqrt[3]{8}\right)}^5}=\frac{2^6\cdot 2:2}{{\left(2^{-6}\right)}^{\frac{3}{4}}\cdot 2^5}=$  

$=\frac{2^7:2}{2^{-\frac{9}{2}}\cdot 2^5}=\frac{2^6}{2^{-4\frac{1}{2}}\cdot 2^5}=\frac{2^6}{2^{\frac{1}{2}}}=2^{5\frac{1}{2}}=2^{\frac{11}{2}}$  

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$\text{b)}\frac{12\sqrt{{32}^2}-{\left(-\frac{1}{2}\right)}^{-8}}{2^{\frac{1}{3}}\cdot 4^{\frac{1}{3}}+{\left(6^{\frac{1}{2}}\cdot 3^{-\frac{1}{2}}\right)}^2}=\frac{12\cdot 32-{\left(-2\right)}^8}{{\left(2\cdot 4\right)}^{\frac{1}{3}}+{\left(6^{\frac{1}{2}}\right)}^2\cdot {\left(3^{-\frac{1}{2}}\right)}^2}=$  

$=\frac{3\cdot 4\cdot 2^5-2^8}{8^{\frac{1}{3}}+6\cdot 3^{-1}}=\frac{3\cdot 2^2\cdot 2^5-2\cdot 2^7}{{\left(2^3\right)}^{\frac{1}{3}}+6\cdot \frac{1}{3}}=\frac{3\cdot 2^7-2\cdot 2^7}{2+2}=$  

$=\frac{\left(3-2\right)\cdot 2^7}{4}=\frac{2^7}{2^2}=2^5$