Tenth Philippine Mathematical Olympiad

Solution:
$\frac{22}{21}$

Note that
$$2 \cos x-4 \sin x=\sqrt{20}\left(\frac{2}{\sqrt{20}} \cos x-\frac{4}{\sqrt{20}} \sin x\right)$$
Let $\varphi$ be a real number such that $\cos \varphi=\frac{2}{\sqrt{20}}$ and $\sin \varphi=\frac{4}{\sqrt{20}}$. We obtain
$$2 \cos x-4 \sin x=\sqrt{20} \cos (x+\varphi)$$
Then
$$0 \leq (2 \cos x-4 \sin x)^{2}=20 \cos ^{2}(x+\varphi) \leq 20$$
Note here that we can particularly choose a value of $x$ so that $20 \cos ^{2}(x+\varphi)=20$, and a value of $x$ so that $20 \cos ^{2}(x+\varphi)=0$. Furthermore, we get
$$1 \leq 1+(2 \cos x-4 \sin x)^{2} \leq 21$$
and so
$$\frac{1}{21} \leq \frac{1}{1+(2 \cos x-4 \sin x)^{2}} \leq 1$$
Thus, the sum of the maximum and minimum values of the given expression is $1+\frac{1}{21}=\frac{22}{21}$.