I know I can use the following: $$\mathcal{L}^{-1}\{e^{-as}F(s)\} = u(t-a)f(t-a)$$ $$\mathcal{L}^{-1}\{\frac{n!}{s^{n+1}}\} = t^n$$ $$\mathcal{L}^{-1}\{F(s-a)\} = e^{at}f(t)$$
but I'm confused as how to use them. In particular, for the first inverse above, if $a$ is negative, does that mean the equation becomes $u(t+a)f(t+a)$, or does the equation stay the same if we use the problem asked? If we have $e^{-3s}\frac{1}{(s-1)^2}$, why would it be
$$u(t-3)(t-3)e^{t-3}$$
as opposed to
$$u(t+3)(t+3)e^{t+3}$$
If, in this problem, the $a$ is negative?
$$$$
A related problem. Note that, the Laplace transform of $ f(t)= t e^t $ is
$$ F(s) = \frac{1}{(s-1)^2} $$
Now, using the fact
$$ \mathcal{L}^{-1}\{e^{-as}F(s)\} = u(t-a)f(t-a), $$
we have
$$ \mathcal{L}^{-1}\{e^{-3s}\frac{1}{(s-1)^2}\} = u(t-3)(t-3)e^{t-3} $$
Note:
To find the Laplace transform of $x^n g(x)$, one can use the following property
$$ \mathcal{L}(x^ng(x))=(-1)^n \frac{d^n}{ds^n} G(s), $$
where $G(s)$ is the Laplace transform of $g(x)$. For instance in your case you have the function $ t e^t $, then its Laplace transform is
$$ (-1) \frac{d}{ds}\frac{1}{s-1}=\frac{1}{(s-1)^2}. $$