(i)

$F\left(s\right)=\dfrac{s\left(s+3\right)\left(s+5\right)}{\left(s+1\right)\left(s+4\right)} $

$F(s)=\dfrac{s\left[s^2+8s+15\right]}{s^2+5s+4}=\dfrac{s^3+8s^2+15s}{s^2+5s+4}=\dfrac{N\left(s\right)}{D\left(s\right)} $

$poles\ :s=-1,\ s=-4 $

$zeros\ :s=0\ \ s=-3,\ s=-5 $

Therefore all poles & zeros  lie on the left half of s-plane as shown :-

-No poles on the jw axis

$-\ Even\ part\ of\ numerator\ of\ F\left(s\right)=m_1=8s^2 $

$Odd\ part\ of\ numerator\ of\ F\left(s\right)=n_1=s^3+15s $

$Even\ part\ of\ denominator\ of\ F\left(s\right)=m_2\left(s\right)=s^2+4 $

$Odd\ part\ of\ denominator\ of\ F\left(s\right)=n_2\left(s\right)=5s $

$\begin{align*} A\left({\omega{}}^2\right)&=\ \ m_1m_2-n_1n_2\ \vert{}\ s=j\omega{} \\ &=8s^2\left(s^2+4\right)-\left(s^3+15s\right)5s\vert{}\ s=j\omega{} \\ &=3s^4-43s^2\vert{}s=j\omega{} \\ &=3{\omega{}}^4+43{\omega{}}^2 \end{align*}$

$\therefore{}A\left({\omega{}}^2\right)\geq{}0\ for\ all\ \omega{}\geq{}0 $

$-\ \ F\left(s\right)\ is\ real\ for\ s\ real\ $

As all condition are satisfied, F(s) is a positive real function.

(ii) $F\left(s\right)=\dfrac{2s^2+2s+4}{\left(s+1\right)\left(s^2+2\right)}=\dfrac{2s^2+2s+4}{\left(s+1\right)(s+j\sqrt{2)}\ (s-j\sqrt{2)}}$

$Let\ N\left(s\right)=2s^2+2s+4 $

Performing Hurwitz test

$even\ part\ of\ N\left(s\right)=2s^2+4 $

$odd\ part\ of\ N\left(s\right)=2s $

As all equation terms are positive N(s) is Hurwitz

This all zeros lie on the left half of the s-plane

$poles\ \ s=-1,\pm{}j\sqrt{2}\ this\ ,all\ poles\ lie\ on\ the\ left\ half\ of\ the\ s-plane\ $

$F\left(s\right)=\dfrac{\dfrac{4}{3}}{s+1}+\dfrac{\dfrac{1}{1-j\sqrt{2}}}{s+j\sqrt{2}}+\dfrac{\dfrac{1}{1+j\sqrt{2}}}{s-j\sqrt{2}} $

As the residue corresponding to poles on the jw axis are not purely led F(s) is not a positive real function.