All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(a)
$w^2 = (a+bi)^2 = a^2 - b^2 + 2abi$

$\frac{w^2}{w^*} = \frac{a^2 - b^2 + 2abi}{a-bi} \times \frac{a+bi}{a+bi}$

$= \frac{1}{a^2 + b^2}(a^3 - 3ab^2 + 3a^2bi - b^3 i)$

For $\frac{w^2}{w^*}$ to be purely imaginary, $Re{\frac{w^2}{w^*}} = 0$

$\Rightarrow a^3 - 3ab^2 = 0$

$a = 0 \mathrm{(NA)}~ \mathrm{or}~ a^2 - 3b^2 = 0$

$b= \pm \frac{a}{\sqrt{3}}$

$\therefore w = a \pm \frac{a}{\sqrt{3}}i$

(bi)
$z^5 = -32i = 2^5 e^{-i\frac{\pi}{2}}$

$z^5 = 2^5 e^{i(-\frac{\pi}{2}+2k\pi)}$ for $k = 0, \pm 1, \pm 2$

$z = 2e^{i(\frac{-\pi + 4k \pi}{10})}$

$\Rightarrow |z| = 2$

$\mathrm{arg}(z) = - \frac{\pi}{10}, - \frac{\pi}{2}, - \frac{9 \pi}{10}, \frac{3 \pi}{10}, \frac{7 \pi}{10}$

(bii)
$z_1 = 2e^{i\frac{7\pi}{10}}$ and $z_2 = 2e^{i\frac{-9\pi}{10}}$

$\angle OZ_2Z_2 = \frac{1}{2}(\pi - \frac{2\pi}{5}) = \frac{3\pi}{10}$

$\mathrm{arg}(Z_1 - Z_2) = \frac{3\pi}{10} + \frac{\pi}{10} = \frac{2\pi}{5}$

By Cosine Rule,

$|Z_1 - Z_2|^2 = 2^2 + 2^2 - 2(2)(2)\mathrm{cos} (\frac{2\pi}{5})$

$|Z_1 - Z_2|^2 = 8 - 8\mathrm{cos} (\frac{2\pi}{5})$

$|Z_1 - Z_2|^2 = 8[1 - \mathrm{cos} (\frac{2\pi}{5})]$

$|Z_1 - Z_2|^2 = 8[2\mathrm{sin^2} (\frac{\pi}{5})]$

$|Z_1 - Z_2|^2 = 16\mathrm{sin^2} (\frac{\pi}{5})$

$|Z_1 - Z_2| = 4\mathrm{sin} (\frac{\pi}{5})$