\(=\dfrac{\left(\sqrt{7}-3\right)\left(\sqrt{7}+1\right)}{6}+\sqrt{\dfrac{4}{8-2\sqrt{7}}}\\ =\dfrac{4-2\sqrt{7}}{6}+\dfrac{2}{\sqrt{7}-1}=\dfrac{4-2\sqrt{7}}{6}+\dfrac{2\left(\sqrt{7}-1\right)}{6}\\ =\dfrac{4-2\sqrt{7}+2\sqrt{7}-2}{6}=\dfrac{2}{6}=\dfrac{1}{3}\)

