$$\tan^{-1}\frac{1+\sqrt{3}}{3+\sqrt{3}} + \sec^{-1}\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}} = $$

First term: $$\frac{1+\sqrt{3}}{3+\sqrt{3}} = \frac{1+\sqrt{3}}{\sqrt{3}(\sqrt{3}+1)} = \frac{1}{\sqrt{3}}$$. So $$\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$$.

Second term: $$\frac{8+4\sqrt{3}}{6+3\sqrt{3}} = \frac{4(2+\sqrt{3})}{3(2+\sqrt{3})} = \frac{4}{3}$$. So $$\sec^{-1}\sqrt{\frac{4}{3}} = \sec^{-1}\frac{2}{\sqrt{3}} = \cos^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{6}$$.

Sum = $$\frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}$$.