The general algorithm to solve these sorts of problems is to expand each side as an infinite series, and the compute enough terms that the difference between values is smaller than the omitted terms.
Good formulas for $\pi$ are difficult to prove, and so usually do not appear in introductory courses. The best technique (IMHO) is the first few terms in the continued fraction expansion of $\pi$: $$3<\frac{333}{106}<\pi<\frac{355}{113}<\frac{22}{7} \tag{1}$$ To verify these yourself, pick your favorite series or physical experiment to determine $\pi$ and compute until you have enough digits to apply the Babylonian algorithm for continued fractions; the convergents necessarily bound $\pi$ from either side.
On the other hand, the standard series defining $e$ is already very good. So our goal is going to try to manipulate the expression so that we have something that depends only slowly on $\pi$, but might be sensitive to $e$.
First, simplify: $$2\ln{\!\left(\frac{\sqrt{\pi}}{2\sqrt{\frac{1}{5}}}\right)}=\ln{\!\left(\frac{5\pi}{4}\right)}$$ Now exponentiate and multiply by $4/5$; we want to show $$\pi<\frac{4}{5}e^{1+\frac{1}{e}}$$ Well, \begin{align*}
\frac{4}{5}e^{1+\frac{1}{e}}&=4e\cdot\exp{(e^{-1})} \\
&=\frac{4}{5}\left(\sum_{k=0}^{\infty}{\frac{1}{k!}}\right)\sum_{k=0}^{\infty}{\frac{e^{-k}}{k!}} \\
&=\frac{4}{5}\left(\sum_{k=0}^{\infty}{\frac{1}{k!}}\right)\sum_{k=0}^{\infty}{\frac{1}{k!}\sum_{j=0}^{\infty}{\frac{(-k)^j}{j!}}} \\
&>\frac{4}{5}\left(\sum_{k=0}^K{\frac{1}{k!}}\right)\sum_{k=0}^K{\frac{1}{k!}\sum_{j=0}^{2\left\lfloor\frac{K^2}{4}\right\rfloor+1}{\frac{(-k)^j}{j!}}} \tag{2}
\end{align*} where $K$ is an arbitrary cutoff. (The cutoff for $j$ probably seems a bit mysterious. It needs to be an odd index so that we have an underestimate, and the general rule of thumb is that an outer power series should have a cutoff that is "little-o" of the inner cutoff.)
Now, compute out (2) until it is larger than a continued fraction on the RHS of (1). As it turns out, the first few terms of (2) are $$\frac{4}{5},\frac{8}{5},\frac{7}{3},\frac{124}{45},\frac{20\,288\,671}{6\,531\,840},\frac{338\,258\,747\,357}{107\,775\,360\,000},\frac{1\,086\,051\,649\,968\,809\,811\,589}{345\,728\,180\,109\,312\,000\,000},\frac{368\,396\,503\,404\,331\,527\,428\,943\,031\,481}{117\,264\,747\,927\,582\,254\,039\,040\,000\,000},\frac{138\,588\,995\,483\,500\,733\,892\,865\,856\,936\,548\,310\,244\,500\,953}{44\,114\,032\,165\,202\,403\,200\,910\,993\,380\,710\,809\,600\,000\,000},\dots$$ I cheated and used Mathematica to compute these fractions. But you could do this by hand — it's just tedious.
In any case, the last fraction I listed above is greater than $\frac{355}{113}$. So
Thus \begin{align*}
\frac{4}{5}e^{1+\frac{1}{e}}&>\frac{4}{5}\left(\sum_{k=0}^8{\frac{1}{k!}}\right)\sum_{k=0}^8{\frac{1}{k!}\sum_{j=0}^{33}{\frac{(-k)^j}{j!}}} \\
&=\frac{138\,588\,995\,483\,500\,733\,892\,865\,856\,936\,548\,310\,244\,500\,953}{44\,114\,032\,165\,202\,403\,200\,910\,993\,380\,710\,809\,600\,000\,000} \\
&>\frac{355}{133} \\
&>\pi
\end{align*}
