Solution: Recall that the Laplace transformation of a function $f$ is given by \[ F(s)\coloneqq \mathcal{L} \left( f(t) \right) = \int_0^{\infty} f(t)e^{-{st} }~\mathrm{d} t. \] We will use the following results for solving the problems.

1. $\mathcal{L} \left( f(t)+g(t) \right) = F(s) + G(s)$.
2. $\mathcal{L} \left( \alpha \cdot f(t) \right) = \alpha \cdot F(s)$.
3. $\mathcal{L} \left( \sin \omega t \right) = \frac{\omega }{s^2 + \omega ^2} $.
4. $\mathcal{L} \left( \sinh \omega t \right) = \frac{\omega }{s^2 - \omega ^2}$.

Now observe that \begin{align*} \mathcal{L} \left( 2\sin 3t + 4\sinh 3t \right) & = \mathcal{L} (2 \sin 3t) + \mathcal{L} (4\sinh 3t) \\ & = 2\cdot\mathcal{L} (\sin 3t) + 4\cdot \mathcal{L} (\sinh 3t) \\ & = 2\cdot\frac{3}{s^2 + 3^2} + 4\cdot \frac{3}{s^2 - 3^2} \\ & = \frac{6}{s^2 + 9} + \frac{12}{s^2 - 9} \\ & = \frac{6s^2 -54 + 12s^2 +108}{\left( s^2 + 9 \right) \left( s^2 -9 \right) } \\ & = \frac{18 s^2 + 54}{s^4 - 81}. \end{align*} Thus, \[ \textcolor{blue}{\boxed{\mathcal{L} \left( 2\sin 3t + 4\sinh 3t \right) = \frac{18s^2 + 54}{s^4 - 81}} } \]