$$\cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=$$

If $$4 \sin ^{-1} x+6 \cos ^{-1} x=3 \pi$$, where $$-1 \leq x \leq 1$$, then $$x=$$

$$\left[\sin \left(\tan ^{-1} \frac{3}{4}\right)\right]^2+\left[\sin \left(\tan ^{-1} \frac{4}{3}\right)\right]^2=$$

The value of $$\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)$$ is