$\text{a)}$  

$x={\left(\frac{\sqrt{3}}{2}\right)}^{-4}={\left(\frac{2}{\sqrt{3}}\right)}^4=\frac{2^4}{{\sqrt{3}}^4}=\frac{16}{9}=1\frac{7}{9}=1,\left(7\right)$  

$y=\sqrt{2}\cdot \sqrt[3]{2}\cdot \sqrt[6]{2}=2^{\frac{1}{2}}\cdot 2^{\frac{1}{3}}\cdot 2^{\frac{1}{6}}=2^{\frac{1}{2}+\frac{1}{3}+\frac{1}{6}}=2^{\frac{3}{6}+\frac{2}{6}+\frac{1}{6}}=2^{\frac{6}{6}}=2^1=2>1,78$  

 

Zatem otrzymujemy, że  $x<y$  

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$\text{b)}$  

$x={\left(\frac{2^6}{6^3}\right)}^{-\frac{2}{3}}={\left(\frac{6^3}{2^6}\right)}^{\frac{2}{3}}={\left(\frac{6^3}{4^3}\right)}^{\frac{2}{3}}={\left({\left(\frac{6}{4}\right)}^3\right)}^{\frac{2}{3}}={\left(\frac{6}{4}\right)}^2={\left(\frac{3}{2}\right)}^2=\frac{9}{4}=2\frac{1}{4}=2,25$  

$y={\left(\sqrt[3]{1,6}\right)}^6={1,6}^{\frac{6}{3}}={1,6}^2=1,6\cdot 1,6=2,56$  

 

Zatem otrzymujemy, że  $x<y$