Answer:  1)  $cosht\int _0^te^usinhu$

Now, let us find $L(sinhu)=\dfrac{1}{s^2-1}$

$\therefore L(e^usinhu)=\dfrac{1}{(s-1)^2-1}$

$\therefore L(e^usinhu)=\dfrac{1}{s(s-2)}$

$\therefore L[\int _0^t e^usinhu \space du]=\dfrac{1}{s}\cdot\dfrac{1}{s(s-2)}$

$\therefore L[\int _0^t e^usinhu \space du]=\dfrac{1}{s^2(s-2)}$

$\therefore L[cosht \cdot \int _0^t e^usinhu \space du]=L[\dfrac{e^t+e^{-t}}{2}\int_0^te^usinhudu]$

$=\dfrac{1}{2}L[ e^t \int_0^t e^usinhudu]+\dfrac{1}{2}L[e^{-t}\int _0^te^usinhudu]$

$=\dfrac{1}{2}[\dfrac{1}{(s-1)^2(s-1-2)}+\dfrac{1}{(s+1)^2(s+1-2)}]$

$L[cosht\int _0^te^usinhu]=\dfrac{1}{2}[\dfrac{1}{(s-1)^2(s-3)}+\dfrac{1}{(s+1)^2(s-1)}]$

2) $t\sqrt{1+sint}$

$\sqrt{1+sint}=\sqrt{sin^2(\dfrac{t}{2})+cos^2(\dfrac{t}{2})+2sint(\dfrac{t}{2})cos(\dfrac{t}{2})}$

$t\sqrt{1+sint}=sin(\dfrac{t}{2})+cos(\dfrac{t}{2})$

Taking laplace on both sides , 

$L[t\sqrt{1+sint}]=L[sin(\dfrac{t}{2})+cos(\dfrac{t}{2})]$

$=\dfrac{\dfrac{1}{2}}{s^2+(\dfrac{1}{2})^2}+\dfrac{s}{s^2+(\dfrac{1}{2})^2}$

$=\dfrac{1}{2} \cdot \dfrac{4}{4s^2+1} +\dfrac{4s}{4s^2+1}$

$\therefore L\sqrt{1+sint}=\dfrac{2(2s+1)}{4s^2+1}$

$\therefore L[t\sqrt{1+sint}]=-\dfrac{d}{ds}[\dfrac{2(2s+1)}{4s^2+1}]$

$=-2[\dfrac{(4s^2+1)2-(2s+1)8s}{(4s^2+1)^2}]$

$=-2\dfrac{[-8s^2-8s+2]}{(4s^2+1)^2}$

$\therefore L[t\sqrt{1+sint}]=4\dfrac{4s^2+4s-1}{(4s^2+1)^2}$