$a)\frac{{\left(2^3\right)}^4\cdot {\left(2^7\right)}^4}{{\left(2^5\right)}^4:{\left(2^3\right)}^4}=\frac{2^{12}\cdot 2^{28}}{2^{20}:2^{12}}=\frac{2^{12+28}}{2^{20-12}}=\frac{2^{40}}{2^8}=2^{32}$  

$b)\frac{2^5\cdot 4^3\cdot {64}^4}{8^2\cdot {128}^3}=\frac{2^5\cdot {\left(2^2\right)}^3\cdot {\left(2^6\right)}^4}{{\left(2^3\right)}^2\cdot {\left(2^7\right)}^3}=$   $\frac{2^5\cdot 2^6\cdot 2^{24}}{2^6\cdot 2^{21}}=\frac{2^{5+6+24}}{2^{6+21}}=\frac{2^{35}}{2^{27}}=2^{35-27}=2^8$  

$c)\frac{{81}^3:3^3}{{27}^5:9^5}=\frac{{\left(3^4\right)}^3:3^3}{{\left(3^3\right)}^5:{\left(3^2\right)}^5}=\frac{3^{12}:3^3}{3^{15}:3^{10}}=\frac{3^9}{3^5}=3^{9-4}=3^5$  

$d)\frac{{\left({\left(0,2\right)}^5\right)}^2\cdot {\left(0,008\right)}^4}{{\left(0,04\right)}^8:0,0016}=\frac{{\left(0,2\right)}^{10}\cdot {\left({\left(0,2\right)}^3\right)}^4}{{\left({\left(0,2\right)}^2\right)}^8:{\left(0,2\right)}^4}=\frac{{\left(0,2\right)}^{10}\cdot {\left(0,2\right)}^{12}}{{\left(0,2\right)}^{16}:{\left(0,2\right)}^4}=\frac{{\left(0,2\right)}^{22}}{{\left(0,2\right)}^{12}}={\left(0,2\right)}^{22-12}={\left(0,2\right)}^{10}$