$\text{a)}\frac{2}{3\sqrt{6}}=\frac{2\cdot \sqrt{6}}{3\sqrt{6}\cdot \sqrt{6}}=\frac{2\sqrt{6}}{3\cdot 6}=\frac{2\sqrt{6}}{18}=\frac{\sqrt{6}}{9}$  

$\text{b)}\frac{2+\sqrt{2}}{\sqrt{2}}=\frac{\left(2+\sqrt{2}\right)\cdot \sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}=\frac{2\sqrt{2}+2}{2}=\frac{2\left(\sqrt{2}+1\right)}{2}=\sqrt{2}+1$  

$\text{c)}\frac{3-2\sqrt{3}}{3\sqrt{3}}=\frac{\left(3-2\sqrt{3}\right)\cdot \sqrt{3}}{3\sqrt{3}\cdot \sqrt{3}}=\frac{3\sqrt{3}-2\cdot 3}{3\cdot 3}=\frac{3\sqrt{3}-6}{9}=\frac{3\left(\sqrt{3}-2\right)}{9}=\frac{\sqrt{3}-2}{3}$  

$\text{d)}\frac{2}{2-\sqrt{2}}=\frac{2\cdot \left(2+\sqrt{2}\right)}{\left(2-\sqrt{2}\right)\cdot \left(2+\sqrt{2}\right)}=\frac{2\left(2+\sqrt{2}\right)}{4-2}=\frac{2\left(2+\sqrt{2}\right)}{2}=2+\sqrt{2}$  

$\text{e)}\frac{3}{\sqrt{3}-\sqrt{2}}=\frac{3\cdot \left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\cdot \left(\sqrt{3}+\sqrt{2}\right)}=\frac{3\left(\sqrt{3}+\sqrt{2}\right)}{3-2}=3\left(\sqrt{3}+\sqrt{2}\right)=3\sqrt{3}+3\sqrt{2}$  

$\text{f)}\frac{5\sqrt{3}}{\sqrt{5}+2}=\frac{5\sqrt{3}\cdot \left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\cdot \left(\sqrt{5}-2\right)}=\frac{5\sqrt{3}\left(\sqrt{5}-2\right)}{5-4}=5\sqrt{3}\left(\sqrt{5}-2\right)=5\sqrt{15}-10\sqrt{3}$  

$\text{g)}\frac{2\sqrt{2}}{\sqrt{2}-\sqrt{3}}=\frac{2\sqrt{2}\cdot \left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\cdot \left(\sqrt{2}+\sqrt{3}\right)}=\frac{2\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}{2-3}=\frac{2\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}{-1}=-2\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)=-4-2\sqrt{6}$  

$\text{h)}\frac{2\sqrt{3}}{3\sqrt{2}+\sqrt{3}}=\frac{2\sqrt{3}\cdot \left(3\sqrt{2}-\sqrt{3}\right)}{\left(3\sqrt{2}+\sqrt{3}\right)\cdot \left(3\sqrt{2}-\sqrt{3}\right)}=\frac{2\sqrt{3}\left(3\sqrt{2}-\sqrt{3}\right)}{18-3}=\frac{6\sqrt{6}-6}{15}=\frac{6\left(\sqrt{6}-1\right)}{15}=\frac{2\left(\sqrt{6}-1\right)}{5}=\frac{2\sqrt{6}-2}{5}$