Given $\frac{1}{4} \bigg[ sin^{-1}\frac{2\sqrt 2}{3}+sin^{-1}\frac{1}{3} \bigg]$
We know that $( sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt{1-y^2}+y\sqrt{1-x^2} \bigg])$
By taking $(x=\frac{2\sqrt 2}{3}\:and\:y=\frac{1}{3})$
$(\sqrt{1-y^2}=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3})$
$(\sqrt{1-x^2}=\sqrt{1-\frac{8}{9}}=\frac{1}{3})$
$( \Rightarrow) ( \frac{1}{4} \bigg[ sin^{-1} \bigg( \frac{2\sqrt 2}{3} \sqrt{1-\frac{1}{9}} +\frac{1}{3} \sqrt{1-\frac{8}{9}} \bigg) \bigg])$
$=\frac{1}{4}sin^{-1}\bigg(\frac{2\sqrt{2}}{3}.\frac{2\sqrt{2}}{3}+\frac{1}{3}.\frac{1}{3}\bigg))$
$ =( \frac{1}{4}sin^{-1} \bigg(\frac{8}{9}+\frac{1}{9}\bigg) )$
$ \frac{1}{4}sin^{-1}1=\frac{1}{4}.\frac{\pi}{2}=\frac{\pi}{8}$