Consider the function
\[F(s) = \frac{2}{(s-4)(s+7)}.\]We do a partial fraction decomposition (which is almost always a good first step). That is we write
\[\frac{2}{(s-4)(s+7)} = \frac{A}{s-4}+\frac{B}{s+7}.\]Multiplying by the denominator on the LHS gives
\begin{equation}\label{eqn:1} 2 = A(s+7) + B(s-4). \end{equation}
Then setting $s = 4$ gives
\[2 = A(4+7) + B(4-4) = 11A\\ A = \frac{2}{11}.\]We can also look at the coefficient of $s$ in \eqref{eqn:1} gives
\[0s = (A+B)s\\ B = -A = -\frac{2}{11}.\]So \(F(s) = \frac{2}{11} \left(\frac{1}{s-4}- \frac{1}{s+7} \right).\)
Hence
\[\mathcal{L}^{-1}(F) = \frac{2}{11} \mathcal{L}^{-1}\left(\frac{1}{s-4} \right) - \frac{2}{11} \mathcal{L}^{-1}\left(\frac{1}{s+7} \right)\\ = \frac{2}{11} \left( e^{4t} - e^{-7t}\right).\]Consider the function
\[F(s) = \frac{24s}{(s+1)(s^2+8s+31)}.\]Again, we start with partial fractions:
\[\frac{24s}{(s+1)(s^2+8s+31)} = \frac{A}{s+1} + \frac{Bs+C}{s^2+8s+31},\]because the term $s^2+8s+31$ cannot be factored.
Multiplying by the denominator on the left hand side gives:
\begin{equation} \label{eqn:2} 24s = A(s^{2} +8s+31) + (Bs+C)(s+1). \end{equation}
Setting $s = -1$ gives
\[-24 = A((-1)^2 +8(-1)+31) + (B(-1)+C)(-1+1) = 24 A\\ A = -1.\]Looking at the coefficient of $s^2$ on both sides of \eqref{eqn:2} gives
\[0s^2 = (A+B)s^2\\ B = -A = 1.\]Lastly, setting $s = 0$ in \eqref{eqn:2} gives
\[0 = 31 A + C\\ C = -31 A = 31,\]and therefore
\begin{equation} \label{eqn:3} F(s) = \frac{24}{(s+1)(s^2+8s+31) } = -\frac{1}{s+1} + \frac{s+31}{s^2+8s+31}. \end{equation}
We now complete the square on the $s^2+8s+31$ term. This gives
\[s^2+8s+31 = (s-a)^2+b^2\\ s^2+8s+31 = s^2-2as+a^2+b^2\\ -2a = 8\\ a^2 +b^2 = 31,\]which in turn gives
\[s^2+8s+31 = (s+4)^2 + \sqrt{15}^2.\]Going back to \eqref{eqn:3}, we have
\begin{equation} \label{eqn:4} F(s) = -\frac{1}{s+1} + \frac{s+31}{(s+4)^2 + \sqrt{15}^2}. \end{equation}
We want to make the last term on the right-hand side of equation \eqref{eqn:4} be of the form
\[\frac{\tilde{A}(s+4) + \tilde{B}(\sqrt{15})}{(s+4)^2 + \sqrt{15}^2},\]in order to use the Laplace transform table.
We do this by adding zero in a clever way, $s = (s+4-4)$, and multiplying by 1 in a clever way, $1 = \sqrt{15}/\sqrt{15}$ so that \eqref{eqn:4} becomes:
\[F(s) = -\frac{1}{s+1} + \frac{s+31}{(s+4)^2 + \sqrt{15}^2}\\ = -\frac{1}{s+1} + \frac{(s+4-4)+31}{(s+4)^2+\sqrt{15}^2}\\ = -\frac{1}{s+1} + \frac{(s+4) + -4 +31}{(s+4)^2+\sqrt{15}^2}\\ = -\frac{1}{s+1} + \frac{s+4}{(s+4)^2+\sqrt{15}^2} + \frac{27}{(s+4)^2+\sqrt{15}^2}\\ = -\frac{1}{s+1} + \frac{s+4}{(s+4)^2+\sqrt{15}^2} + \frac{27}{\sqrt{15}}\frac{\sqrt{15}}{(s+4)^2+\sqrt{15}^2}.\]This is much easier to take the inverse Laplace transform of:
\[\mathcal{L}^{-1}(F) = -\mathcal{L}^{-1}\left(\frac{1}{s+1}\right) + \mathcal{L}^{-1}\left(\frac{s+4}{(s+4)^2+\sqrt{15}^2} \right) + \frac{27}{\sqrt{15}}\mathcal{L}^{-1}\left(\frac{\sqrt{15}}{(s+4)^2+\sqrt{15}^2} \right)\\ = -e^{-t} + e^{-4t}\cos(\sqrt{15}t) + \frac{27}{\sqrt{15}} e^{-4t}\sin(\sqrt{15}t).\]Consider the function
\[F(s) = \frac{se^{-2s}}{s^2+4s+13} + \frac{e^{-\pi s}}{s+7}.\]Whenever we see an inverse Laplace transform problem that has an $e^{-cs}$ term in it we factor out that term. That is we re-write
\[F(s) = e^{-2s} \left(\frac{s}{s^2+4s+13}\right) + e^{-\pi s}\left( \frac{1}{s+7}\right).\\ = e^{-2s} G(s) + e^{-\pi s} H(s),\]where
\[G(s) = \frac{s}{s^2+4s+13} = \frac{s}{(s+2)^2+3^2}\\ = \frac{(s+2-2)}{(s+2)^2+3^2} \\ = \frac{s+2}{(s+2)^2+3^2} - \frac{2}{3} \frac{3}{(s+2)^2+3^2}\]and
\[H(s) = \frac{1}{s+7}.\]Then
\[g(t) = \mathcal{L}^{-1}\left(G(s) \right) = \mathcal{L}^{-1} \left( \frac{s+2}{(s+2)^2+3^2} - \frac{2}{3} \frac{3}{(s+2)^2+3^2}\right)\\ = e^{-2t}\cos(3t)+ \frac{2}{3} e^{-2t} \sin(3t),\\ h(t) = \mathcal{L}^{-1}\left(H(s)\right) = e^{-7t}.\]Therefore, by equation (12) in the table, we can write
\[\mathcal{L}^{-1}(e^{-2s}G(s)) = g(t-2) u_2\\ \mathcal{L}^{-1}(e^{-\pi s}) = h(t-\pi) u_\pi.\]Therefore,
\[\mathcal{L}^{-1} \left( \frac{se^{-2s}}{s^2+4s+13} + \frac{e^{-\pi s}}{s+7}\right)= \mathcal{L}^{-1}\left(e^{-2s} G(s) + e^{-\pi s} H(s)\right)\\ = g(t-2)u_2 + h(t-\pi)u_\pi,\]where $g$ and $h$ are as defined above.