Answer
(a) $(-4\sqrt 2,-4\sqrt 2)$
(b) $(4\sqrt 3, \frac{5\pi}{6})$ and $(-4\sqrt 3, \frac{11\pi}{6})$
Work Step by Step
(a) Use the conversion equations $x=r\cdot cos\theta, y=r\cdot sin\theta$ , we have $x=8cos\frac{5\pi}{4}=-8\times\frac{\sqrt 2}{2}=-4\sqrt 2$ and $y=8sin\frac{5\pi}{4}=-8\times\frac{\sqrt 2}{2}=-4\sqrt 2$ so the answer is $(-4\sqrt 2,-4\sqrt 2)$
(b) Use the conversion formula $|r|=\sqrt {x^2+y^2}=\sqrt {(-6)^2+(2\sqrt 3)^2}=\sqrt {48}=4\sqrt 3$ and $\theta=tan^{-1}(-\frac{2\sqrt 3}{6})=tan^{-1}(-\frac{\sqrt 3}{3})$ within $0\leq\theta\lt 2\pi$, we get $\theta=\frac{5\pi}{6}, \frac{11\pi}{6}$ and the final answers are $(4\sqrt 3, \frac{5\pi}{6})$ and $(-4\sqrt 3, \frac{11\pi}{6})$
