a) Obliczamy

$5^{-4}={\left(\frac{1}{5}\right)}^4=\frac{1}{625}$  

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b) Obliczamy

${\left(-\frac{2}{5}\right)}^{-3}={\left(-\frac{5}{2}\right)}^3=-\frac{125}{8}$

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c) Obliczamy

${0,01}^{-2}={\left(\frac{1}{100}\right)}^{-2}={100}^2=10000$  

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d) Obliczamy

${\left(\sqrt{2}\right)}^{-4}={\left(\frac{1}{\sqrt{2}}\right)}^4=\frac{1}{{\left(\sqrt{2}\right)}^4}=\frac{1}{4}$  

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e) Obliczamy 

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

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f) Obliczamy 

${\left(\sqrt{5}-\sqrt{2}\right)}^{-2}={\left(\frac{1}{\sqrt{5}-\sqrt{2}}\right)}^2$  

Usuwamy niewymierność z mianownika ułamka

$\frac{1}{\sqrt{5}-\sqrt{2}}\cdot \frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}+\sqrt{2}}{5-2}=\frac{\sqrt{5}+\sqrt{2}}{3}$  

Korzystając ze wzoru skróconego mnożenia na kwadrat sumy, otrzymujemy

${\left(\frac{1}{\sqrt{5}-\sqrt{2}}\right)}^2={\left(\frac{\sqrt{5}+\sqrt{2}}{3}\right)}^2=\frac{5+2\cdot \sqrt{5}\cdot \sqrt{2}+2}{9}=\frac{7+2\sqrt{10}}{9}$