$a)$     $a={\left(3^{-1}+3\right)}^{-1}+3^{-1}={\left(\frac{1}{3}+3\right)}^{-1}+3^{-1}={\left(3\frac{1}{3}\right)}^{-1}+3^{-1}={\left(\frac{10}{3}\right)}^{-1}+3^{-1}=$

$=\frac{3}{10}+\frac{1}{3}=\frac{9}{30}+\frac{10}{30}=\frac{19}{30}$   

$b={\left\{{\left[{\left(\frac{1}{3}\right)}^{-1}+{\left(\frac{1}{3}\right)}^{-1}\right]}^{-1}\right\}}^{-1}={\left[{\left(3+3\right)}^{-1}\right]}^{-1}={\left(6^{-1}\right)}^{-1}={\left(\frac{1}{6}\right)}^{-1}=6$   

$a<b$   

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$b)$     $a={\left[{\left(6^{-1}+6^{-1}\right)}^{-1}-{\left(1\frac{1}{5}\right)}^{-1}\right]}^{-1}={\left[{\left(\frac{1}{6}+\frac{1}{6}\right)}^{-1}-{\left(\frac{6}{5}\right)}^{-1}\right]}^{-1}=$

$={\left[{\left(\frac{2}{6}\right)}^{-1}-\frac{5}{6}\right]}^{-1}={\left[{\left(\frac{1}{3}\right)}^{-1}-\frac{5}{6}\right]}^{-1}={\left(3-\frac{5}{6}\right)}^{-1}={\left(\frac{18}{6}-\frac{5}{6}\right)}^{-1}=$

$={\left(\frac{13}{6}\right)}^{-1}=\frac{6}{13}$    

$b={\left\{{\left[{\left(\frac{5}{6}\right)}^{-1}+{\left(\frac{5}{6}\right)}^{-1}\right]}^{-1}-6^{-1}\right\}}^{-1}={\left[{\left(\frac{6}{5}+\frac{6}{5}\right)}^{-1}-6^{-1}\right]}^{-1}=$   

$={\left[{\left(\frac{12}{5}\right)}^{-1}-6^{-1}\right]}^{-1}={\left(\frac{5}{12}-\frac{1}{6}\right)}^{-1}={\left(\frac{5}{12}-\frac{2}{12}\right)}^{-1}={\left(\frac{3}{12}\right)}^{-1}=$

$=\frac{12}{3}=4$   

$a<b$

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$c)$     $a={\left(\frac{9\sqrt[3]{9}}{\sqrt[5]{27}}\right)}^4={\left(\frac{9\cdot 9^{\frac{1}{3}}}{{27}^{\frac{1}{5}}}\right)}^4={\left(\frac{9^1\cdot 9^{\frac{1}{3}}}{{\left(3^3\right)}^{\frac{1}{5}}}\right)}^4={\left(\frac{9^{1+\frac{1}{3}}}{3^{3\cdot \frac{1}{5}}}\right)}^4={\left(\frac{9^{1\frac{1}{3}}}{3^{\frac{3}{5}}}\right)}^4=$  

$={\left(\frac{9^{\frac{4}{3}}}{3^{\frac{3}{5}}}\right)}^4={\left(\frac{{\left(3^2\right)}^{\frac{4}{3}}}{3^{\frac{3}{5}}}\right)}^4={\left(\frac{3^{2\cdot \frac{4}{3}}}{3^{\frac{3}{5}}}\right)}^4={\left(\frac{3^{\frac{8}{3}}}{3^{\frac{3}{5}}}\right)}^4={\left(3^{\frac{8}{3}-\frac{3}{5}}\right)}^4=$  

$={\left(3^{\frac{40}{15}-\frac{9}{15}}\right)}^4={\left(3^{\frac{31}{15}}\right)}^4=3^{\frac{31}{15}\cdot 4}=3^{\frac{124}{15}}=3^{8\frac{4}{15}}$  

$b={\left(\sqrt[3]{{\left(\sqrt[5]{81}\right)}^4}\right)}^{10}={\left({\left({\left(\sqrt[5]{81}\right)}^4\right)}^{\frac{1}{3}}\right)}^{10}={\left({\left(\sqrt[5]{81}\right)}^4\right)}^{\frac{1}{3}\cdot 10}={\left({\left(\sqrt[5]{81}\right)}^4\right)}^{\frac{10}{3}}=$   

$={\left(\sqrt[5]{81}\right)}^{4\cdot \frac{10}{3}}={\left(\sqrt[5]{81}\right)}^{\frac{40}{3}}={\left({81}^{\frac{1}{5}}\right)}^{\frac{40}{3}}=\left({81}^{\frac{1}{5}\cdot \frac{40}{3}}\right)={81}^{\frac{40}{15}}={\left(3^4\right)}^{\frac{40}{15}}=$  

$=3^{4\cdot \frac{40}{15}}=3^{\frac{160}{15}}=3^{\frac{32}{3}}=3^{10\frac{2}{3}}$  

$a<b$

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$d)$     $a=\frac{2\sqrt{2}}{\sqrt[3]{2}}=\frac{2\cdot 2^{\frac{1}{2}}}{2^{\frac{1}{3}}}=\frac{2^1\cdot 2^{\frac{1}{2}}}{2^{\frac{1}{3}}}=\frac{2^{1+\frac{1}{2}}}{2^{\frac{1}{3}}}=\frac{2^{\frac{3}{2}}}{2^{\frac{1}{3}}}=2^{\frac{3}{2}-\frac{1}{3}}=2^{\frac{9}{6}-\frac{2}{6}}=2^{\frac{7}{6}}=2^{1\frac{1}{6}}$  

$b=\sqrt[5]{\frac{64}{\sqrt{2}}}={\left(\frac{64}{\sqrt{2}}\right)}^{\frac{1}{5}}={\left(\frac{2^6}{2^{\frac{1}{2}}}\right)}^{\frac{1}{5}}=\frac{{\left(2^6\right)}^{\frac{1}{5}}}{{\left(2^{\frac{1}{2}}\right)}^{\frac{1}{5}}}=\frac{2^{6\cdot \frac{1}{5}}}{2^{\frac{1}{2}\cdot \frac{1}{5}}}=\frac{2^{\frac{6}{5}}}{2^{\frac{1}{10}}}=2^{\frac{6}{5}-\frac{1}{10}}=$

$=2^{\frac{12}{10}-\frac{1}{10}}=2^{\frac{11}{10}}=2^{1\frac{1}{10}}$    

$a>b$