$a\text{)}\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1$

  $b\text{)}\frac{8}{3-\sqrt{5}}=\frac{8\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}=$ $\frac{8\left(3+\sqrt{5}\right)}{9-5}=\frac{8\left(3+\sqrt{5}\right)}{4}=2\left(3+\sqrt{5}\right)$

  $c\text{)}\frac{6}{3-2\sqrt{3}}=\frac{6\left(3+2\sqrt{3}\right)}{\left(3-2\sqrt{3}\right)\left(3+2\sqrt{3}\right)}=$ $\frac{6\left(3+2\sqrt{3}\right)}{9-4\cdot 3}=\frac{6\left(3+2\sqrt{3}\right)}{9-12}=$ $\frac{6\left(3+2\sqrt{3}\right)}{-3}=-2\left(3+2\sqrt{3}\right)$

  $d\text{)}\frac{3}{\sqrt{13}+\sqrt{10}}=$ $\frac{3\left(\sqrt{13}-\sqrt{10}\right)}{\left(\sqrt{13}+\sqrt{10}\right)\left(\sqrt{13}-\sqrt{10}\right)}$ $=\frac{3\left(\sqrt{13}-\sqrt{10}\right)}{13-10}=\frac{3\left(\sqrt{13}-\sqrt{10}\right)}{3}=\sqrt{13}-\sqrt{10}$