The value of $\frac{9}{4} \sin^{-1}\frac{2\sqrt{2}}{3} + \frac{9}{4} \sin^{-1}\frac{1}{3}$ is :

$\frac{9 \pi}{8}$

$\frac{9}{4} \sin^{-1}\frac{2\sqrt{2}}{3} + \frac{9}{4} \sin^{-1}\frac{1}{3}$

we know that for triangle

$A = \sin^{-1} \frac{2\sqrt{2}}{3}$      $B = \sin^{-1}\frac{1}{3}$

so  $(A + B) = \frac{\pi}{2}  ⇒  \sin^{-1}\frac{2\sqrt{2}}{3} + \sin^{-1}\frac{1}{3} = \frac{\pi}{2}$

$\frac{9}{4} \sin^{-1}\frac{2\sqrt{2}}{3} + \frac{9}{4} \sin^{-1}\frac{1}{3} = \frac{9 \pi}{8}$