$\text{a)}x={81}^{20}={\left(3^4\right)}^{20}=3^{80}$  

$y={1024}^8={\left(2^{10}\right)}^8=2^{80}$  

$3^{80}>2^{80},$  bo  $3>2,$  więc

$x>y$  

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$\text{b)}x={128}^{36}={\left(2^7\right)}^{36}=2^{252}$  

$y=4^{125}={\left(2^2\right)}^{125}=2^{250}$  

$2^{252}>2^{250},$  bo  $252>250,$  więc

$x>y$  

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$\text{c)}x={\left(\frac{1}{27}\right)}^{-48}={\left(3^{-3}\right)}^{-48}=3^{144}$  

$y={243}^{28}={\left(3^5\right)}^{28}=3^{144}$  

$3^{148}>3^{140},$  bo  $148>144,$  więc

$x>y$  

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$\text{d)}x={\left(2\sqrt{2}\right)}^{28}={\left(2^{\frac{3}{2}}\right)}^{28}=2^{3\cdot 14}=2^{42}$  

$y={\left(\frac{\sqrt{8}}{64}\right)}^{-6}={\left(\frac{\sqrt{2^3}}{2^6}\right)}^{-6}={\left(\frac{2^{\frac{3}{2}}}{2^{\frac{12}{2}}}\right)}^{-6}={\left(2^{-\frac{9}{2}}\right)}^{-6}=2^{9\cdot 3}=2^{27}$  

$2^{42}>2^{27},$  bo  $42>27,$  więc

$x>y$