Q2 might still benefit from an explicit answer in terms of a special function.

The inverse Laplace transform of ${\rm sech}(\sqrt{2\lambda})=1/\cosh(\sqrt{2\lambda})$ follows from an entry in Table 2 in <A HREF="https://www.sciencedirect.com/science/article/pii/0022072888870013">Theta functions; transform tables and examples for electrochemists:</A>

$$\int_0^\infty dt\, e^{-\lambda t}\,\frac{\partial}{\partial x}\theta_1\left(\frac{x}{2c}\biggl|\frac{i\pi t}{c^2}\right)=\frac{c\cosh(x\sqrt{\lambda})}{\cosh(c\sqrt{\lambda})},\;\;|x|<c, $$
where $\theta_1$ is a <A HREF="https://en.wikipedia.org/wiki/Theta_function#Jacobi_theta_function">Jacobi theta function</A>.   
The inverse Laplace transform of $1/\cosh(\sqrt{2\lambda})$ is thus given by
$${\cal L}^{-1}\left(\frac{1}{\cosh(\sqrt{2\lambda})}\right)=\frac{1}{\sqrt 2}\lim_{x\rightarrow 0}\frac{\partial}{\partial x}\theta_1\left(\frac{x}{2\sqrt{2}}\biggl|\frac{i\pi t}{2}\right).$$