$ \displaystyle i)\ Let\ \\[2ex] p(s)=\ 2\ {\tan h}^{-1}s \\[2ex] \displaystyle =2\times{}\frac{1\ }{2}\log{\left(\frac{1+s}{1-s}\right)} \\[2ex] \displaystyle =log(s+1)-log(1-s) \\[2ex] \displaystyle \therefore{}p'(s)=\ \frac{1}{s+1}-\frac{1}{1-s}\bullet{}-1\\[2ex] \ \ \ \displaystyle p^{'}(s)=\frac{1}{s+1}-\frac{1}{s-1} \\[2ex] $

$\displaystyle Using\ Formula, \\[2ex] \displaystyle L^{-1}\left[F\left(s\right)\right]=\ -\frac{1}{t}\ L^{-1}[F^{'}\left(s\right)] \\[2ex] \displaystyle \therefore{}L^{-1}\left[2\ {\tan h}^{-1}s\ \ \right]=-\frac{1}{t}L^{-1}[\frac{1}{s+1}-\frac{1}{s-1}] \\[2ex] \displaystyle =-\frac{1}{t}(e^{-t}-e^t) \\[2ex] \displaystyle =\frac{1}{t}(e^t-e^{-t}) \\[2ex] \displaystyle =\frac{1}{t}\times{}2\sin ht \\[2ex] \displaystyle =\frac{2}{t}\sin ht \\[2ex]$

$\displaystyle ii)\frac{s^2}{{(s}^2+1)(s^2+4\ )}\\[2ex] \displaystyle =\ \frac{s^2+1-1}{{(s}^2+1)(s^2+4\ )} \\[2ex] \displaystyle =\ \frac{s^2+1\ }{{(s}^2+1)(s^2+4\ )}-\frac{1}{3}\frac{3}{{(s}^2+1)(s^2+4\ )} \\[2ex] \displaystyle =\frac{1}{s^2+4\ }-\frac{1}{3}\frac{4-1}{{(s}^2+1)\left(s^2+4\ \right)} \\[2ex] \displaystyle =\frac{1}{s^2+4\ }-\frac{1}{3}\frac{\left(s^2+\ 4\right)-\left(s^2+1\right)}{{(s}^2+1)\left(s^2+4\ \right)} \\[2ex] \displaystyle =\frac{1}{s^2+4\ }-\frac{1}{3}\left[\frac{1}{s^2+1\ }-\frac{1}{s^2+4\ }\right] \\[2ex] \displaystyle =\ -\frac{1}{3}\frac{1}{s^2+1\ }+\left(1+\frac{1}{3}\ \right)\frac{1}{s^2+4\ } \\[2ex] \displaystyle =\ -\frac{1}{3}(\frac{1}{s^2+1\ })+\frac{4}{3}(\frac{1}{s^2+4\ }) \\[2ex] $

$\displaystyle L^{-1}\left[\ -\frac{1}{3}\left(\frac{1}{s^2+1\ }\right)+\frac{4}{3}\left(\frac{1}{s^2+4\ }\right)\right]\\[2ex] =\ -\frac{1}{3}\ L^{-1}\left(\frac{1}{s^2+1\ }\right)+\frac{4}{3}L^{-1}\left(\frac{1}{s^2+4\ }\right) \\[2ex] \displaystyle =\ -\frac{\sin t}{3}+\frac{2}{3}\sin{2t\ }\ \ \\[2ex] $