$\text{a)}{64}^{\frac{2}{3}}+{64}^{\frac{3}{2}}={\left(\sqrt[3]{64}\right)}^2+{\left(\sqrt{64}\right)}^3=4^2+8^3=16+512=528$  

$\text{b)}{\left(2\frac{7}{9}\right)}^{\frac{1}{2}}+{\left(1\frac{9}{16}\right)}^{\frac{1}{2}}={\left(\frac{25}{9}\right)}^{\frac{1}{2}}+{\left(\frac{25}{16}\right)}^{\frac{1}{2}}=\sqrt{\frac{25}{9}}+\sqrt{\frac{25}{16}}=\frac{5}{3}+\frac{5}{4}=\frac{20}{12}+\frac{15}{12}=\frac{35}{12}$  

$\text{c)}{0,04}^{-\frac{1}{2}}\cdot {81}^{-\frac{1}{2}}={\left(\sqrt{0,04}\right)}^{-1}\cdot {\left(\sqrt{81}\right)}^{-1}={0,2}^{-1}\cdot 9^{-1}={\left(\frac{1}{5}\right)}^{-1}\cdot \frac{1}{9}=5\cdot \frac{1}{9}=\frac{5}{9}$  

$\text{d)}{625}^{\frac{3}{4}}:{125}^{\frac{2}{3}}={\left(\sqrt[4]{625}\right)}^3:{\left(\sqrt[3]{125}\right)}^2=5^3:5^2=5$  

$\text{e)}{27}^{1\frac{1}{3}}\cdot 9^{-\frac{3}{2}}+{32}^{\frac{2}{5}}={27}^{\frac{4}{3}}\cdot {\left(\sqrt{9}\right)}^{-3}+{\left(\sqrt[5]{32}\right)}^2={\left(\sqrt[3]{27}\right)}^4\cdot 3^{-3}+2^2=3^4\cdot \frac{1}{3^3}+4=3+4=7$  

$\text{f)}{\left(3\frac{3}{8}\right)}^{-\frac{2}{3}}+{25}^{\frac{1}{2}}\cdot {81}^{-\frac{1}{2}}={\left(\frac{27}{8}\right)}^{-\frac{2}{3}}+\sqrt{25}\cdot {\left(\sqrt{81}\right)}^{-1}={\left(\sqrt[3]{\frac{27}{8}}\right)}^{-2}+5\cdot 9^{-1}={\left(\frac{3}{2}\right)}^{-2}+5\cdot \frac{1}{9}={\left(\frac{2}{3}\right)}^2+\frac{5}{9}=\frac{4}{9}+\frac{5}{9}=\frac{9}{9}=1$