The expression $$\frac{2(\sqrt{2} + \sqrt{6})}{3\sqrt{2 + \sqrt{3}}}$$ is equal to 

$$\frac{2(\sqrt{2} + \sqrt{6})}{3\sqrt{2 + \sqrt{3}}}$$

Let's rationalize the denominator.

$$\frac{2(\sqrt{2}+\sqrt{6})}{3\sqrt{2+\sqrt{3}}}\cdot\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2-\sqrt{3}}}$$
$$\frac{2(\sqrt{2}+\sqrt{6})}{3\sqrt{1}}\cdot\sqrt{2-\sqrt{3}}=\frac{2}{3}\left(\sqrt{\ 2}+\sqrt{\ 6}\right)\cdot\sqrt{2-\sqrt{3}}$$
Let's assume that the whole as t.
Now $$t^2$$ = $$\frac{4}{9}\left(\sqrt{\ 2}+\sqrt{\ 6}\right)^2\cdot2-\sqrt{3}$$
$$\frac{4}{9}\left(2+6+2\sqrt{\ 12}\right)\cdot\left(2-\sqrt{3}\right)=\frac{4}{9}\left(8+4\sqrt{\ 3}\right)\left(2-\sqrt{\ 3}\right)=\frac{16}{9}\left(2+\sqrt{\ 3}\right)\left(2-\sqrt{\ 3}\right)=\frac{16}{9}$$
$$t^2$$=$$\frac{16}{9}$$
t=4/3.