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

We need to evaluate $$\tan^{-1}\frac{1+\sqrt{3}}{3+\sqrt{3}} + \sec^{-1}\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}}$$.

Observe that $$\frac{1+\sqrt{3}}{3+\sqrt{3}} = \frac{1+\sqrt{3}}{\sqrt{3}(\sqrt{3}+1)} = \frac{1}{\sqrt{3}},$$ so $$\tan^{-1}\frac{1}{\sqrt{3}} = \frac{\pi}{6}$$.

Similarly, $$\frac{8+4\sqrt{3}}{6+3\sqrt{3}} = \frac{4(2+\sqrt{3})}{3(2+\sqrt{3})} = \frac{4}{3},$$ giving $$\sec^{-1}\sqrt{\frac{4}{3}} = \sec^{-1}\frac{2}{\sqrt{3}}$$. Since $$\sec\theta = \frac{2}{\sqrt{3}}$$ implies $$\cos\theta = \frac{\sqrt{3}}{2},$$ it follows that $$\theta = \frac{\pi}{6}$$.

Adding these two values yields $$\frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}$$.

The answer is Option 3: $$\frac{\pi}{3}$$.