(1) If $L^{-1} [f(s)] = f(t)$

$dL^{-1} \bigg[\frac{d^n}{ds^n} f(s) \bigg] = (-1)^n t^n f(t)$ ..............(A)

$f(s) = log\bigg(\frac{s^2 + 4}{s^2 + 9} \bigg)$

Then $f(s) = log(s^2 + 4) - log(s^2 + 9)$

Now $\frac{d}{ds} f(s) = \frac{1}{s^2 + 4} (2s) - \frac{1}{s^2 + 9} (2s)$

$\therefore \frac{d}{ds} f(s) = \frac{2s}{s^2 + 4} - \frac{2s}{s^2 + 9}$

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

$L^{-1} \bigg[\frac{d}{ds} f(s) \bigg] = 2cost2t - 2cos3t$

Now from eq(A), we can write

(-1)tf(t) = 2cos2t - 2cos3t

$\therefore f(t) = \frac{2cos3t - 2cos2t}{t}$

(2) Let, $f(s) = \frac{s}{(s^2 + 4)(s^2 + 9)}$

By using partial fraction method,

$\frac{s}{(s^2 + 4)(s^2 + 9)} = \frac{As + B}{s^2 + 4} + \frac{Cs + D}{s^2 + 9}$

$\therefore s = (As + B)(s^2 + 9) + (Cs + D)(s^2 + 4)$

Now, equating 'degree 1' coefficients.

sA + 4c = 1

and now equating 'degree 3' coefficients

A + C = 0 $\therefore A = -C$

Hence, C = $ - \frac{1}{5}$

-9C + 4C = 1

Hence, A = $\frac{1}{5}$

Now equating 'degree 2' & 'degree 0' coefficients we get, B + D = 0 B = -D

9B + 4D = 0 $\therefore$ D = 0 hence B = 0

$A = \frac{1}{5}, B = 0, C = - \frac{1}{5}, D = 0$

$\therefore \frac{s}{(s^2 + 4)(s^2 + 9)} = \frac{1}{5} \frac{s}{s^2 + 4} - \frac{1}{5} \frac{s}{s^2 + 9}$

$\therefore \frac{s}{(s^2 + 4)(s^2 + 9)} = \frac{1}{5} (cos2t - cos3t)$

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