(1) We will first find $cos2t cos3t = \frac{1}{2} [cos5t + cos t]$

$\therefore L[cos2t cos3t] = L\bigg[\frac{1}{2}(cos5t + cost)\bigg]$

$\therefore L[cos2t cos3t] = \frac{1}{2}L[cos5t] + \frac{1}{2}L[cost]$

$\therefore L[cos2t cos3t] = \frac{1}{2} \bigg[\frac{s}{s^2 + 25} + \frac{s}{s^2 + 1} \bigg]$

$\therefore L[e^{-3t} cos2t cos3t] = \frac{1}{2}\bigg[\frac{(s + 3)}{(s +3)^2 + 25} + \frac{s + 3}{(s + 3)^2 + 1} \bigg]$

$ L[e^{-3t} cos2t cos3t] = (-1)\frac{d}{ds} \bigg[\frac{s + 3}{s^2 + 6s + 34} + \frac{s + 3}{s^2 + 6s + 10}\bigg]$

$\therefore L[e^{-3t} cos2t cos3t] = (-1) \frac{1}{2}\bigg[\frac{(s^2 + 6s + 34)(1) - (s + 3)(2s + 6)}{(s^2 + 6s + 34)^2} + \frac{(s^2 + 6s + 10)(1) - (s + 3)(2s + 6)}{(s^2 + 6s + 10)^2}\bigg] -\frac{1}{2}\bigg[\frac{(s^2 + 6s + 34) - (2s^2 + 12s + 18)}{(s^2 + 6s + 34)^2} + \frac{(s^2 + 6s + 10)(1) - (2s^2 + 12s + 18)}{(s^2 + 6s + 10)^2}\bigg]$

$\therefore L[e^{-3t} cos2t cos3t] = \frac{-1}{2} \bigg[\frac{-s^2 -6s + 16}{(s^2 + 6s + 34)^2} + \frac{(-s^2 -6s - 8)}{(s^2 + 6s + 10)^2}\bigg]$

$\therefore L[e^{-3t} cos2t cos3t] = \frac{1}{2}\bigg[\frac{s^2 + 6s - 16}{(s^2 + 6s + 34)^2} + \frac{(s^2 + 6s + 8)}{(s^2 + 6s + 10)^2}\bigg]$

$\therefore L[te^{-3t} cos2t cos3t] = \frac{1}{2}\bigg[\frac{s^2 + 6s - 16}{(s^2 + 6s + 34)^2} + \frac{(s^2 + 6s + 8)}{(s^2 + 6s + 10)^2}\bigg]$

(2) Here f(t) = $\frac{sin3t}{t}$

We will first find $L[sin(3t)] = \frac{3}{s^2 + 3^2} = \frac{3}{s^2 + 9}$

$\therefore L\bigg[\frac{sin3t}{t}\bigg] = \int\limits_{s}^{\infty} \frac{3}{s^2 + 9} ds = 3*\frac{1}{3} \bigg[tan^{-1}\bigg(\frac{s}{3}\bigg)\bigg]_{s}^{\infty}$

$\therefore L\bigg[\frac{sin3t}{t}\bigg] = \frac{\pi}{2} - tan^{-1} \bigg(\frac{s}{3}\bigg) = cot^{-1}\bigg(\frac{s}{3}\bigg) = f(s)$

$\therefore L\bigg[\frac{d}{dt}\bigg(\frac{sin3t}{t}\bigg)\bigg] = sf(s) - f(0) = scot^{-1} \bigg(\frac{s}{3}\bigg) - \frac{\pi}{2}$

$\therefore L\bigg[\frac{d}{dt}\bigg(\frac{sin3t}{t}\bigg)\bigg] = scot^{-1}\bigg(\frac{s}{3}\bigg) - \frac{\pi}{2}$