$\text{a)}a={\left(2\sqrt{2}\right)}^{4\sqrt{3}}={\left(\sqrt{8}\right)}^{4\sqrt{3}}={\left(8^{\frac{1}{2}}\right)}^{4\sqrt{3}}=8^{\frac{1}{2}\cdot 4\sqrt{3}}=8^{2\sqrt{3}}$  

$b^{-1}={\left({\left(\frac{1}{8}\right)}^{2\sqrt{3}}\right)}^{-1}={\left(\frac{1}{8}\right)}^{-2\sqrt{3}}=8^{2\sqrt{3}}$

A więc liczba a jest odwrotna do b.

$a=b^{-1}$  

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$\text{b)}a={36}^{1+\sqrt{2}}={\left(6^2\right)}^{1+\sqrt{2}}=6^{2+2\sqrt{2}}=6^{2+\sqrt{8}}$  

$b^{-1}={\left(6^{1-\sqrt{8}}\right)}^{-1}=6^{\sqrt{8}-1}$  

A więc liczba a nie jest odwrotna do b.

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$\text{c)}{a}^{-1}={\left({25}^{2-3\sqrt{5}}\right)}^{-1}={\left(5^2\right)}^{3\sqrt{5}-2}=5^{6\sqrt{5}-4}=5^{6\sqrt{5}}\cdot 5^{-4}=5^{\sqrt{36\cdot 5}}\cdot \frac{1}{625}=0,0016\cdot 5^{\sqrt{180}}=b$  

A więc liczba a jest odwrotna do b.

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$\text{d)}{b}^{-1}={\left({\left(\frac{36}{100}\right)}^{1+\sqrt{0,75}}\right)}^{-1}={\left(\frac{100}{36}\right)}^{1+\sqrt{0,75}}={\left(\frac{25}{9}\right)}^{1+\sqrt{0,75}}={\left({\left(\frac{5}{3}\right)}^2\right)}^{1+\sqrt{0,75}}={\left(\frac{5}{3}\right)}^{2+2\sqrt{0,75}}={\left(1\frac{2}{3}\right)}^{2+\sqrt{4\cdot 0,75}}={\left(1\frac{2}{3}\right)}^{2+\sqrt{3}}=a$  

A więc liczba a jest odwrotna do b.