Let's do some calculus! (61)

$\begin{aligned} L & = \lim_{x \to \infty} \left(\frac {\cosh \frac \pi x}{\cos \frac \pi x}\right)^{x^2} & \small \color{#3D99F6} \text{A }1^\infty \text{ cases: }\lim_{x \to a} (f(x))^{g(x)} = e^{\lim_{x\to a}g(x)(f(x)-1)} \\ & = \exp \left(\lim_{x \to \infty} x^2 \left(\frac {\cosh \frac \pi x}{\cos \frac \pi x} - 1\right) \right) & \small \color{#3D99F6} \text{Let }u = \frac 1x \\ & = \exp \left(\lim_{u \to 0} \frac {\cosh (\pi u) - \cos (\pi u)}{u^2\cos \pi u} \right) & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \exp \left(\lim_{u \to 0} \frac {\pi \sinh (\pi u) + \pi \sin (\pi u)}{2u\cos \pi u-u^2\pi \sin \pi u} \right) & \small \color{#3D99F6} \text{Differentiate up and down, and a 0/0 case again.} \\ & = \exp \left(\lim_{u \to 0} \frac {\pi^2 \cosh (\pi u) + \pi^2 \cos (\pi u)}{2\cos \pi u-2u\sin \pi u -2u\pi \sin \pi u-u^2 \pi^2 \cos \pi u} \right) & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }u \text{ again.} \\ & = e^{\pi^2} \end{aligned}$

Therefore, $a=\boxed{2}$ .