$\displaystyle \left(i\right)\ L\left[t\ e^{-3t}\cos{2t}.\cos{3t}\right] \\[2ex] \displaystyle \cos{2t}\cos{3t}=\frac{1}{2}\left(\cos{\left(2t+3t\right)}+\cos{\left(3t-2t\right)}\right)=\frac{1}{2}(\cos{5t}+\cos t)\ \\[2ex] \displaystyle \therefore{},\ L\left[t\ e^{-3t}\cos{2t}.\cos{3t}\right]=\frac{1}{2}\ \left(L\left[t\ e^{-3t}\cos{5t}\right]+L\ \left[t\ e^{-3t}\cos t\right]\right) \\[2ex] $

$\displaystyle \ L\left[\cos{at}\right]=\frac{s}{s^2+a^2} \\[2ex] \begin{align*} \displaystyle L\left[t\cos{5t}\right]&=-\frac{d}{ds}\left[\frac{s}{s^2+25}\right]\\[2ex]&=-\left(\frac{s^2+25-2s\left(s\right)}{{\left(s^2+25\right)}^2}\right)\\[2ex] &=\frac{s^2-25}{{\left(s^2+25\right)}^2} \\[2ex] \end{align*} $

$\displaystyle L\left[e^{-3t}t\cos{5t}\right]=\ \frac{({s+3)}^2-25}{{\left(({s+3)}^2+25\right)}^2}\ \\[2ex] \begin{align*} \displaystyle L\left[t\cos t\right]&=\ -\ \frac{d}{ds}\left[\frac{s}{s^2+1^2}\right]\\[2ex] &=-\left(\frac{s^2+1-2s\left(s\right)}{{\left(s^2+1\right)}^2}\right)\\[2ex] &=\frac{s^2-1}{{\left(s^2+1\right)}^2}\ \\[2ex] \end{align*}$

$\displaystyle L\left[e^{-3t}t\cos t\right]=\ \frac{({s+3)}^2-1}{{\left(({s+3)}^2+1\right)}^2} \\[2ex] \displaystyle \therefore{},\ L\left[t\ e^{-3t}\cos{2t}.\cos{3t}\right]=\frac{1}{2}\left[\ \frac{({s+3)}^2-25}{{\left(({s+3)}^2+25\right)}^2}+\frac{({s+3)}^2-1}{{\left(({s+3)}^2+1\right)}^2}\right]\ \\[2ex] $

$\displaystyle \ \left(ii\right)\ \ L\ \left[\frac{d}{dt}f\left(t\right)\right]=s\ L[f(t)]-f\left(0\right) \\[2ex] $

$\displaystyle L\left[ \frac{d}{dt}\ \left[\frac{\sin{3t}}{t}\right]\ \right]=\ s\ L\left[\frac{\sin{3t}}{t}\right]-\lim_{t\rightarrow{}0}{\frac{\sin{3t}}{t}}\ \\[3ex] \displaystyle =\ s\int_s^{\infty{}}\frac{3}{s^2+3^2}ds-3\lim_{3t\rightarrow{}0}{\frac{\sin{3t}}{3t}} \\[2ex] $

$\displaystyle =\frac{3s}{3}{\left[{\tan}^{-1}{\left(\frac{s}{3}\right)}\right]}_s^{\infty{}}-3 \\[2ex] \displaystyle =s\ \left[\frac{\pi{}}{2}-{ \tan}^{-1}{\left(\frac{s}{3}\right)}\ \right]-3 \\[2ex] \displaystyle =s\ \left[{\cot}^{-1}{\left(\frac{s}{3}\right)}\right]-3 \\[2ex] \displaystyle =s\ {\tan}^{-1}{\ \left(\frac{3}{s}\right)\ }-3\ \ \\[2ex] $