Answer
(a) $\frac{\sqrt 5}{3}$
(b) $\frac{4\sqrt 2}{9}$
Work Step by Step
(a) Let $u=arcsin\frac{2}{3}$, we have $sin(u)=\frac{2}{3}$. Form a right triangle of sides $2, \sqrt 5, 3$ with angle $u$ facing the side $2$, we have $cos(u)=\frac{\sqrt 5}{3}$. Thus $cos(arcsin\frac{2}{3})=cos(u)=\frac{\sqrt 5}{3}$
(b) Let $u=cos^{-1}\frac{1}{3}$, we have $cos(u)=\frac{1}{3}$. Form a right triangle of sides $2\sqrt 2,1,3$ with angle $u$ facing the side $2\sqrt 2$, we have $sin(u)=\frac{2\sqrt 2}{3}$. Thus $sin(2cos^{-1}\frac{1}{3})=sin(2u)=2sin(u)cos(u)=2(\frac{2\sqrt 2}{3})(\frac{1}{3})=\frac{4\sqrt 2}{9}$
