(1) According to differentiation property

$L^{-1}log\frac{s^2 + 4}{s^2 + 9} = \frac{-1}{t}L^{-1}\frac{d}{ds}[log(s^2 + 4) - log(s^2 + 9)]$

$L^{-1}log\frac{s^2 + 4}{s^2 + 9} = \frac{-1}{t}L^{-1}\bigg[\frac{2s}{s^2 + 4} - \frac{2s}{s^2 + 9}\bigg]$

$L^{-1}log\frac{s^2 + 4}{s^2 + 9} = \frac{-1}{t}[2cos2t - 2cos3t]$

$L^{-1}log\frac{s^2 + 4}{s^2 + 9} = \frac{2}{t}(cos3t - cos2t)$

(2) By convolution theorem

$\phi_{1}(s) = \frac{s}{s^2 + 4}$

$f_{1}(t) = cos2t$

$\phi_{2}(s) = \frac{s}{s^2 + 9}$

$f_{2}(t) = \frac{1}{3}sin3t$

$\therefore$ According to definition

$L^{-1}\bigg[\frac{s}{(s^2 + 4)(s^2 + 9)}\bigg] = \int\limits_{0}^{t}\frac{1}{2}cos2u*cos3(t - u)du$

$L^{-1}\bigg[\frac{s}{(s^2 + 4)(s^2 + 9)}\bigg] = \frac{1}{2}\int\limits_{0}^{t}(sin(2u + 3t - 3u) - sin(2u - 3t + 3u))du$

$L^{-1}\bigg[\frac{s}{(s^2 + 4)(s^2 + 9)}\bigg] = \frac{1}{2}\int\limits_{0}^{t}sin(3t - u) - sin(5u - 3t)du$

$L^{-1}\bigg[\frac{s}{(s^2 + 4)(s^2 + 9)}\bigg] = \frac{1}{2}\bigg[\frac{-cos(3t - u)}{-1} + \frac{cos(5u - 3t)}{5}\bigg]_{0}^{t}$

$L^{-1}\bigg[\frac{s}{(s^2 + 4)(s^2 + 9)}\bigg] = \frac{1}{2}\bigg[cos2t + \frac{cos2t}{5} - cos3t - \frac{cos3t}{5}\bigg]$

$L^{-1}\bigg[\frac{s}{(s^2 + 4)(s^2 + 9)}\bigg] = \frac{1}{2}\bigg[\frac{6cos2t - 6cos3t}{5}\bigg]$

$L^{-1}\bigg[\frac{s}{(s^2 + 4)(s^2 + 9)}\bigg] = \frac{3}{5}[cos2t - cos3t]$