\(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)=\frac{\sqrt{2}.\sqrt{2-\sqrt{3}}.\sqrt{2}\left(\sqrt{3}+1\right)}{\sqrt{2}}=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)