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Answer 1

$1)\sqrt{2+\sqrt{3}}=\sqrt{\frac{2*(2+\sqrt{3})}{2}}=\sqrt{\frac{4+2\sqrt{3}}{2}}=\sqrt{\frac{(\sqrt{3}+1)^{2}}{2}}=\frac{\sqrt{3}+1}{\sqrt{2}}$

Аналогично :

$2)\sqrt{2-\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{2}}$

$3)\frac{2+\sqrt{3}}{\sqrt{2}+\frac{\sqrt{3}+1}{\sqrt{2}}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\frac{\sqrt{3}-1}{\sqrt{2}}}=\frac{\sqrt{2}*(2+\sqrt{3})}{2+\sqrt{3}+1}+\frac{\sqrt{2}*(2-\sqrt{3})}{2-\sqrt{3}+1}=\frac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}+\frac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}=$

$=\frac{2\sqrt{6}-2\sqrt{2}+\sqrt{18}-\sqrt{6}+2\sqrt{6}+2\sqrt{2}-\sqrt{18}-\sqrt{6}}{3^{2}-(\sqrt{3})^{2}}=\frac{2\sqrt{6}}{6}=\frac{\sqrt{6}}{3}$