Given that,

$$ \begin{aligned} z & =\frac{2-i}{(1-2 i)^2}=\frac{2-i}{1+4 i^2-4 i} \\ & =\frac{2-i}{1-4-4 i}=\frac{2-i}{-3-4 i} \\ & =\frac{(2-i)}{-(3+4 i)}=-\left[\frac{(2-i)(3-4 i)}{(3+4 i)(3-4 i)}\right] \\ & =-\left(\frac{6-8 i-3 i+4 i^2}{9+16}\right)=-\frac{(-11 i+2)}{25} \\ & =\frac{-1}{25}(2-11 i) \Rightarrow z=\frac{1}{25}(-2+11 i) \\ \therefore \quad\bar{z} & =\frac{1}{25}(-2-11 i)=\frac{-2}{25}-\frac{11}{25} i \end{aligned}$$

Given that, $$\left|z_1\right|=\left|z_2\right|$$

Let $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$

$$\begin{aligned} \Rightarrow \quad & \left|x_1+i y_1\right| =\left|x_2+i y_2\right| \\ \Rightarrow \quad & x_1^2+y_1^2 =x_2^2+y_2^2 \\ \Rightarrow \quad & x_1^2 =x_2^2, y_1^2=y_2^2 \\ \Rightarrow \quad & x_1 = \pm x_2, y_1= \pm y_2 \\ \Rightarrow \quad & z_1 =x_1+i y_1 \text { or } z_1= \pm x_2 \pm i y_2 \end{aligned}$$

Hence, it is not neccessary that $z_1=z_2$.

Given that, $\quad \frac{\left(a^2+1\right)^2}{2 a-i}=x+i y \Rightarrow \frac{\left(a^2+1\right)^2}{(2 a-i)}=x+i y$

$\begin{aligned} & \Rightarrow \quad \frac{\left(a^2+1\right)^2(2 a+i)}{(2 a-i)(2 a+i)}=x+i y \\ & \Rightarrow \quad \frac{\left(a^2+1\right)^2(2 a+i)}{4 a^2+1}=x+i y\end{aligned}$

$\begin{aligned} \Rightarrow \quad x & =\frac{2 a\left(a^2+1\right)^2}{4 a^2+1} \text { and } y=\frac{\left(a^2+1\right)^2}{4 a^2+1} \\ \therefore \quad x^2+y^2 & =4 a^2\left[\frac{\left(a^2+1\right)^2}{4 a^2+1}\right]^2+\left[\frac{\left(a^2+1\right)^2}{4 a^2+1}\right]^2 \\ & =\frac{\left(4 a^2+1\right)\left(a^2+1\right)^4}{\left(4 a^2+1\right)^2}=\frac{\left(a^2+1\right)^4}{\left(4 a^2+1\right)}\end{aligned}$

Let $$z=r(\cos \theta+i \sin \theta)$$

$$\begin{aligned} \therefore \quad z & =4\left[\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}\right] [\because z=r(\cos \theta+i \sin \theta)]\\ & =4\left[\cos \left(\pi-\frac{\pi}{6}\right)+i \sin \left(\pi-\frac{\pi}{6}\right)\right] \\ & =4\left[-\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right] \\ & =4\left[-\frac{\sqrt{3}}{2}+i \frac{1}{2}\right]=-2 \sqrt{3}+2 i \end{aligned}$$

Given that, $$ \begin{aligned} \left|(1+i) \frac{(2+i)}{(3+i)}\right| & =\left|\frac{\left(2+i+2 i+i^2\right)}{(3+i)}\right|=\left|\frac{2+3 i-1}{3+i}\right| \\ & =\left|\frac{1+3 i}{3+i}\right|=\left|\frac{(1+3 i)(3-i)}{(3+i)(3-i)}\right| \\ & =\left|\frac{3+9 i-i-3 i^2}{9-i^2}\right|=\left|\frac{3+8 i+3}{9+1}\right|=\left|\frac{6+8 i}{10}\right| \\ & =\sqrt{\frac{6^2}{100}+\frac{8^2}{100}}=\sqrt{\frac{36+64}{100}}=\sqrt{\frac{100}{100}}=1 \end{aligned} $$