$\text{a)}$  

Korzystamy ze wzoru na różnicę kwadratów:

$\left(a+b\right)\left(a-b\right)=a^2-b^2$  

Mamy więc:

$ab=\sqrt{\sqrt{5}+\sqrt{3}}\cdot \sqrt{\sqrt{5}-\sqrt{3}}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\sqrt{{\left(\sqrt{5}\right)}^3-{\left(\sqrt{3}\right)}^2}=$  

$=\sqrt{5-3}=\sqrt{2}$  

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$\text{b)}$  

Korzystamy ze następujących wzorów:

${\left(a+b\right)}^2=a^2+2ab+b^2$  

$\left(a+b\right)\left(a-b\right)=a^2-b^2$  

Mamy więc:

${\left(a+b\right)}^2={\left(\sqrt{\sqrt{5}+\sqrt{3}}+\sqrt{\sqrt{5}-\sqrt{3}}\right)}^2=$  

$={\left(\sqrt{\sqrt{5}+\sqrt{3}}\right)}^2+2\cdot \sqrt{\sqrt{5}+\sqrt{3}}\cdot \sqrt{\sqrt{5}-\sqrt{3}}+{\left(\sqrt{\sqrt{5}-\sqrt{3}}\right)}^2=$  

$=\sqrt{5}+\sqrt{3}+2\sqrt{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\sqrt{5}-\sqrt{3}=$  

$=2\sqrt{5}+2\sqrt{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2}=2\sqrt{5}+2\sqrt{5-3}=2\sqrt{5}+2\sqrt{2}$  

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$\text{c)}$  

Korzystamy ze wzoru na różnicę kwadratów:

$\left(a+b\right)\left(a-b\right)=a^2-b^2$  

 

Mamy więc:

$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{{\left(\sqrt{\sqrt{5}+\sqrt{3}}\right)}^2}+\frac{1}{{\left(\sqrt{\sqrt{5}-\sqrt{3}}\right)}^2}=\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{5}-\sqrt{3}}=$  

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

$=\frac{1\cdot \left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{1\cdot \left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}=$  

$=\frac{\sqrt{5}-\sqrt{3}}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2}+\frac{\sqrt{5}+\sqrt{3}}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2}=\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{5}+\sqrt{3}}{5-3}=$  

$=\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{5}+\sqrt{3}}{2}=\left(\sqrt{5}-\sqrt{3}\right)+\sqrt{5}+\sqrt{3}\frac{)}{2}=\frac{2\sqrt{5}}{2}=\sqrt{5}$