$(i)Consider\\[2ex] L\left[1-e^{au}\right]=L\left[1\right]-L\left[e^{au}\right]=\frac{1}{s}-\frac{1}{s-a} \\[2ex] \displaystyle \therefore{}\ L\left[\frac{{1-\ e}^{au}}{u}\right]=\int_s^{\infty{}}(\frac{1}{s}-\frac{1}{s-a})\ ds\ \\[3ex] \{L[\frac{f(t)}{t}]=\int_s^{\infty{}}F(s)ds\} \\[3ex] \displaystyle ={\left[\ln{s}-\ln{\left(s-a\right)}\right]}_s^{\infty{}}\\[2ex] \displaystyle ={\left[-\ln{\frac{s-a}{s\ }}\right]}_s^{\infty{}} \\[3ex] \displaystyle =\lim_{s\rightarrow{}\infty{}}{\left(\ln{\left(1-\frac{a}{s}\right)}\right)}+\ln{\frac{s-a}{s\ }}\\[3ex] \displaystyle=\ln{\left(1-0\right)}\ +\ln{\Bigg(\frac{s-a}{s}} \Bigg) \\[2ex] \displaystyle =\ln{(\frac{s-a}{s}})\\[3ex] \\[2ex] \{ L[\int_0^tf\left(x\right)dx]=\frac{1}{s}L[f(x)] \} \\[2ex] \displaystyle \therefore{}L\left[\int_0^t\frac{{1-\ e}^{au}}{u}\ du\right]=\frac{1}{s}L\left[\frac{{1-\ e}^{au}}{u}\right]=\ \frac{1}{s}.\ln{(\frac{s-a}{s\ }}) \\[2ex] $

${ \displaystyle \left(ii\right)\\[2ex] {\left(t\sin h{2t}\right)}^2=t^2\ \sin h \ }^2{2t\ }\\[2ex] \displaystyle =t^2\ {\left(\frac{e^{2t}-e^{-2t}}{2}\right)}^2\\[2ex] \displaystyle=\frac{t^2}{4}\ (e^{4t}+e^{-4t}-2) \\[2ex] $

$ \{L\left[e^{at}f\left(t\right)\right]=F\left(s-a\right);L\left[1\right]=\frac{1}{s}\} \\[3ex] \displaystyle \therefore{}L(\left[e^{4t}\right])=\frac{1}{s-4} \\[2ex] \displaystyle \therefore{}L\left(\left[e^{-4t}\right]\right)=\frac{1}{s+4} \\[2ex] \begin{align*} \displaystyle \therefore{}L\left[2\right]&=\frac{2}{s} \\[2ex] \displaystyle L\left[\left(e^{4t}+e^{-4t}-2\right)\right]&=\ \frac{1}{s-4}+\frac{1}{s+4}-\frac{2}{s} \\[2ex] \displaystyle L\left[t^2\left(e^{4t}+e^{-4t}-2\right)\right]&={\left(-1\right)}^2\frac{d^2}{ds^2}\left(\frac{1}{s-4}+\frac{1}{s+4}-\frac{2}{s}\right) \\[2ex] \displaystyle \{L\left[x^nf\left(x\right)\right]={\left(-1\right)}^nF^{\left(n\right)}&(s)\} \\[2ex] \displaystyle &=\frac{d}{ds}\left[-\frac{1}{{\left(s-4\right)}^2}\ -\frac{1}{{\left(s+4\right)}^2}+\frac{2}{s^2}\right] \\[2ex] \displaystyle &=\frac{2}{{\left(s-4\right)}^3}+\frac{2}{{\left(s+4\right)}^3}-\frac{4}{s^3} \\[2ex] \displaystyle \therefore{}L\left[{\left(t\sinh{2t}\right)}^2\right]&=\frac{1}{4}L\left[t^2\left(e^{4t}+e^{-4t}-2\right)\right] \\[2ex] \displaystyle &=\ \frac{1}{{2\left(s-4\right)}^3}+\frac{1}{{2\left(s+4\right)}^3}-\frac{1}{s^3} \\[2ex] \end{align*} $