(a) e^4t
(b) 4 e^2t + 3 cosh 4t
(c) 2 sin3t + 4 sinh 3t
(d) t^3 + 2t^2 - 4t + 1
Just apply the formulas! Also note that the Laplace Transform is linear (I did two of them for you):
(a) $\displaystyle \mathcal{L}\left\{e^{4t}\right\}=\frac{1}{s-4}$
(c) $\displaystyle \mathcal{L}\left\{2\sin\left(3t\right)+4\sinh\left (3t\right)\right\}=2\mathcal{L}\left\{\sin\left(3t \right)\right\}+4\mathcal{L}\left\{\sinh\left(3t\r ight)\right\}$ $\displaystyle =2\cdot\frac{3}{s^2+9}+4\cdot\frac{3}{s^2-9}=\frac{6s^2-54+12s^2+108}{s^4-81}=\frac{18\left(s^2+3\right)}{s^4-81}$
If you can't use the short cuts, then apply the definition:
$\displaystyle \mathcal{L}\left\{f\left(t\right)\right\}=\int_0^{ \infty}e^{-st}f\left(t\right)\,dt$
I hope this helps...