2. (15 PTS) Use the accompanying tables of Laplace transforms and properties of Laplace transforms to find the Laplace transform of the function below. Note that an appropriate trigonometric identity may be necessary.
\[
f(t)=e^{-2 t} \sin 2 t+e^{3 t} t^{2}
\]

Step 1 ：Given the function \(f(t)=e^{-2 t} \sin 2 t+e^{3 t} t^{2}\), we are to find its Laplace transform.

Step 2 ：Using the table of Laplace transforms, we know that the Laplace transform of \(e^{at}f(t)\) is \(F(s-a)\) and the Laplace transform of \(t^n f(t)\) is \((-1)^n F^{(n)}(s)\).

Step 3 ：Applying these rules, the Laplace transform of \(e^{-2 t} \sin 2 t\) is \(\frac{2}{(s+2)^2 + 2^2}\) and the Laplace transform of \(e^{3 t} t^{2}\) is \(\frac{2!}{(s-3)^(2+1)}\).

Step 4 ：Adding these two results together, we get the Laplace transform of the function \(f(t)\) as \(F(s) = \frac{2}{(s+2)^2 + 4} + \frac{2}{(s-3)^3}\).

Step 5 ：So, the final answer is \(\boxed{F(s) = \frac{2}{(s+2)^2 + 4} + \frac{2}{(s-3)^3}}\). This is the Laplace transform of the given function.