Assuming H(s) = 1

$\therefore{}G\left(s\right)H\left(s\right)=\dfrac{2(s+0.25)}{s^2\ (s+1)(s+0.5)}=\dfrac{2\times{}0.25}{0.5}\ \ \dfrac{1+45}{s^2\left(1+s\right)\left(1+2s\right)}$

$k=\dfrac{2\times{}0.25}{0.5}=1$

$s^n=s^{-2}$

$\therefore{}G\left(s\right)H\left(s\right)=\dfrac{1+45}{s^2\left(1+s\right)\left(1+2s\right)}\ $

$n=-2$

Various factor of G(s) are

1. $2 \ poles\ at\ origin, straight\ line\ of\ slope - 40dB/dec\\ \ \ \ \ \ \ passing\ through\ intersection\ of \omega{}=1 \ \&\ 0dB.$
2. $Simple\ zero\ \left(1+4s\right),\ T_1=4,\ {\omega{}}_{c1}=\dfrac{1}{T_1}=0.25\\ straight\ line\ of\ slope+\dfrac{20dB}{dec}\ \ for\ \omega{}\geq{}0.25$
3. $simple\ pole\dfrac{1}{2s+1},\ T_2=1,\ {\omega{}}_{c2}\dfrac{1}{\ T_2}=1\\ \therefore{}straight\ line\ of\ slope-\dfrac{20dB}{dec}for\ \omega{}\geq{}1$
4. $simple\ pole\dfrac{1}{2s+},\ T_3=2,\ {\omega{}}_{c3}=\dfrac{1}{T_3}=0.5\\ \therefore{}straight\ line\ of\ slope-20\dfrac{dB}{dec}for\ \omega{}\geq{}0.5\ $

$k=1,\ \ n=-2$

$\begin{align*} Starting\ point&=20\ log_{10}k-20\left(n\right)\ dB\\[2ex] &=20\log\left(1\right)-20\left(-2\right)\\[2ex] &=40 \ dB\\ \end{align*}$

$=20\log\left(1\right)-20\left(-2\right)$

$=40\ dB$

$\begin{align*}Starting\ slope&=\ \ 20n\dfrac{db}{decode} \\[2ex] &=20\ \left(-2\right)\\[2ex] &=-40\frac{dB}{decode} \end{align*}$

$put\ s=j\omega{}$

$G(jw) =(jw)^{-2}\dfrac{1+4jw}{(1+jw)(1+2jw)}$