Solution:

Given this equation,

$ H_a(s)=\frac{4}{(s+3)(s+4)}\\ $

To obtain $H(z)$ using the bilinear transformation in $H_a(s)$, replace s by

$ \frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)\\ $

$ H(z)=\left.\frac{4}{(s+3)(s+4)}\right|_{s=\frac{2}{T} \frac{1-z^{-1}}{1+z^{-1}}}=\left.\frac{4}{(s+3)(s+4)}\right|_{s=4 \frac{1-z^{-1}}{1+z^{-1}}}\\ $

$ \begin{aligned} & =\frac{4}{\left[4\left(\frac{1-z^{-1}}{1+z^{-1}}\right)+3\right]\left[4\left(\frac{1-z^{-1}}{1+z^{-1}}\right)+4\right]} \\\\ & =\frac{4}{\left[\frac{4-4 z^{-1}+3+3 z^{-1}}{1+z^{-1}}\right]\left[\frac{4-4 z^{-1}+4+4 z^{-1}}{1+z^{-1}}\right]} \\\\ & =\frac{4\left(1+z^{-1}\right)^2}{\left(7-z^{-1}\right) 8} \\\\ & =\frac{1}{2} \frac{\left(1+z^{-1}\right)^2}{\left(7-z^{-1}\right)}\\ \end{aligned} $