(ii) $\frac{3}{2}+\frac{3 \sqrt{3}}{2} i$

(iii) 16

(iv) 6

(i) $2 z-1=|w| \Rightarrow z=\frac{|w|+1}{2} \in \mathbb{R}, 2 z^3-5 z^2+2 z \in \mathbb{R}$. Thus for $a \in \mathbb{R}$ $2 z^3-5 z^2+2 z+(a+3) \mathrm{i}=0$
$\Rightarrow 2 z^3-5 z^2+2 z=0$ and $a+3=0$
$\Rightarrow^{z(2 z-1)(z-2)}=0$ and $a=-3$
$z(2 z-1)(z-2)=0 \Rightarrow z=0, \frac{1}{2}, 2$
If $z=0$, then $|w|=2 z-1=-1<0$ which is impossible.
If $z=\frac{1}{2}$, then $|w|=0 \Rightarrow w=0$ which contradicts $w \neq 0$.
Hence $z=2$.

(ii) $|w|=2(2)-1=3$
$w=3\left(\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}\right)=\frac{3}{2}+\frac{3 \sqrt{3}}{2} \mathrm{i}$

(iii) $\begin{array}{l}
\left|w^n\right|>20212021 \\
|w|^n>20212021 \\
3^n>20212021 \\
n>\frac{\lg 20212021}{\lg 3} \approx 15.3
\end{array}$
So least $n$ is 16.

(iv) For $w^k$ to be a positive real number,
$\arg w^k = 0,2\pi, 4\pi, … $ $k \in \mathbb{Z}$ 

$ k\arg w = 0,2\pi, 4\pi, … $ $k \in \mathbb{Z}$ 

$\frac{k \pi}{3}= 0,2\pi, 4\pi, … $ $k \in \mathbb{Z}$ 

$k = 0,6, 12, …  $

So least positive integer $k$ is 6 .