Solution:
We know that,
$
\Omega_c=\frac{2}{T} \tan \frac{\omega_c}{2}\\
$
$
\Omega_c=\frac{2}{T} \tan \frac{0.4 \pi}{2}=\frac{1.453}{T}\\
$
This system response of the digital filter is given by,
$
H(z)=H_a(s) \mid=\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)\\
$
$
=\frac{\Omega_c}{\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)+2 \Omega_c}=\frac{\frac{1.453}{T}}{\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)+2\left(\frac{1.453}{T}\right)}\\
$
$
\begin{aligned}\\
& =\frac{1.453\left(1+z^{-1}\right)}{2\left(1-z^{-1}\right)+2\left(1+z^{-1}\right) 1.453} \\\\
& =\frac{1+z^{-1}}{3.376-0.624 z^{-1}}\\
\end{aligned}\\
$