Answer
a) $24\sqrt 2 [\cos (17\pi/12)+i \sin (17\pi/12)]$
b) $\dfrac{4 \sqrt 2}{3} [\cos (-13\pi/12)+i \sin (-13\pi/12)]$
c) $\dfrac{1}{8}[\cos (-\pi/6)+i \sin (-\pi/6)]$
Work Step by Step
a) Here, we have $z=8[\cos (\dfrac{\pi}{6})+i \sin (\dfrac{\pi}{6})]$ and $w=3 \sqrt 2 [\cos (5\pi/4)+i \sin (5\pi/4)]$
$zw=8[\cos (\pi/6)+i \sin (\dfrac{\pi}{6})] \times 3 \sqrt 2 [\cos (5\pi/4)+i \sin (5\pi/4)]=24\sqrt 2 [\cos (17\pi/12)+i \sin (17\pi/12)]$
b) Here, we have $z=8[\cos (\dfrac{\pi}{6})+i \sin (\dfrac{\pi}{6})]$ and $w=3 \sqrt 2 [\cos (5\pi/4)+i \sin (5\pi/4)]$
$\dfrac{z}{w}=\dfrac{8[\cos (\dfrac{\pi}{6})+i \sin (\dfrac{\pi}{6})]}{3 \sqrt 2 [\cos (5\pi/4)+i \sin (5\pi/4)]}=\dfrac{4 \sqrt 2}{3} [\cos (-13\pi/12)+i \sin (-13\pi/12)]$
c) Here, we have $z=8[\cos (\dfrac{\pi}{6})+i \sin (\dfrac{\pi}{6})]$ and $w=3 \sqrt 2 [\cos (5\pi/4)+i \sin (5\pi/4)]$
$\dfrac{1}{z}=\dfrac{1}{8}[\cos (-\pi/6)+i \sin (-\pi/6)]$