$e)$
= $\frac{\sqrt[]{2}}{2+\sqrt[]{4+2\sqrt[]{3}}}$+ $\frac{\sqrt[]{2}}{2-\sqrt[]{4-2\sqrt[]{3}}}$
= $\frac{\sqrt[]{2}}{2+\sqrt[]{(\sqrt[]{3}+1)^{2}}}$+ $\frac{\sqrt[]{2}}{2-\sqrt[]{(\sqrt[]{3}-1)^{2}}}$
= $\frac{\sqrt[]{2}}{2+\sqrt[]{3}+1}$+ $\frac{\sqrt[]{2}}{2-\sqrt[]{3}+1}$
= $\frac{\sqrt[]{2}}{3+\sqrt[]{3}}$+ $\frac{\sqrt[]{2}}{3-\sqrt[]{3}}$
= $\sqrt[]{2}$.($\frac{3+\sqrt[]{3}+3-\sqrt[]{3}}{(3+\sqrt[]{3})(3-\sqrt[]{3})})$
= $\sqrt[]{2}$. $\frac{6}{9-3}$
= $\sqrt[]{2}$. 1
= $\sqrt[]{2}$
$f)$
= $\frac{(\sqrt[]{5}+2)^{2}-8\sqrt[]{5}}{2\sqrt[]{5}-4}$
= $\frac{5+4+4\sqrt[]{5}-8\sqrt[]{5}}{2(\sqrt[]{5}-2)}$
= $\frac{9-4\sqrt[]{5}}{2(\sqrt[]{5}-2)}$
= $\frac{(\sqrt[]{5}-2)^{2}}{2(\sqrt[]{5}-2)}$
= $\frac{\sqrt[]{5}-2}{2}$
