Решение:
\[\sqrt{108-12\sqrt{3}sin^2(\frac{7\pi}{12})} = \sqrt{108-12\sqrt{3}(\frac{\sqrt{6}+\sqrt{2}}{4})^2}=\sqrt{108-12\sqrt{3}(\frac{6+2+2\sqrt{12}}{16})}=\sqrt{108-12\sqrt{3}(\frac{8+4\sqrt{3}}{16})}=\sqrt{108-3\sqrt{3}(2+\sqrt{3})}=\sqrt{108-6\sqrt{3}-9}=\sqrt{99-6\sqrt{3}}\]