Solution:

Given $H(s)=\frac{2}{(s+1)(s+2)}$

Substitute $s=\frac{2}{T}\left[\frac{1-z^{-1}}{1+z^{-1}}\right]$ in $H(s)$ to get $H(z)$

$ \begin{aligned} H(z) & =\left.H(s)\right|_{s=\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)} \\\\ & =\left.\frac{2}{(s+1)(s+2)}\right|_{s=\frac{2}{T}}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)\\ \end{aligned}\\ $

Given $T=1 \mathrm{sec}$

$ \begin{aligned} H(z) & =\frac{2}{\left\{2\left(\frac{1-z^{-1}}{1+z^{-1}}\right)+1\right\}\left\{2\left(\frac{1-z^{-1}}{1+z^{-1}}\right)+2\right\}} \\\\ & =\frac{2\left(1+z^{-1}\right)^2}{\left(3-z^{-1}\right)(4)} \\\\ & =\frac{\left(1+z^{-1}\right)^2}{6-2 z^{-1}} \\\\ & =\frac{0.166\left(1+z^{-1}\right)^2}{\left(1-0.33 z^{-1}\right)}\\ \end{aligned} $