Just substitute?

$\begin{aligned} \frac{1}{\cos{A}} + \frac{2\cos{A}}{\cos{2A}} & = \frac {\cos{2A} + 2\cos^2{A}} {\cos{A}\cos{2A}} \\ & = \frac {\cos{2A} + 2\cos^2{A} - 1 + 1} {\cos{A}\left(2\cos^2{A}-1\right)} \\ & = \frac {2\cos{2A} + 1} {2\cos^3{A}-\cos{A}} \\ & = \frac {2\cos{2A} + 1} {\frac{1}{2}\left(4\cos^3{A}-3\cos{A}\right) + \frac{3}{2}\cos{A}-\cos{A}} \\ & = \frac {2\cos{2A} + 1} {\frac{1}{2}\cos{3A} + \frac{1}{2}\cos{A}} \end{aligned}$

We know that:

$\begin{aligned} \cos{\frac{\pi}{7}} + \cos{\frac{3\pi}{7}} + \cos{\frac{5\pi}{7}} & = \frac{1}{2} \\ \cos{\frac{\pi}{7}} + \cos{\frac{3\pi}{7}} - \cos{\frac{2\pi}{7}} & = \frac{1}{2} \\ \cos{\frac{\pi}{7}} + \cos{\frac{3\pi}{7}} & = \cos{\frac{2\pi}{7}} + \frac{1}{2} \end{aligned}$

Therefore,

$\begin{aligned} \frac{1}{\cos{\frac{\pi}{7}}} + \frac{2\cos{\frac{\pi}{7}}}{\cos{\frac{2\pi}{7}}} & = \frac {2\cos{\frac{2\pi}{7}} + 1} {\frac{1}{2}\cos{\frac{3\pi}{7}} + \frac{1}{2}\cos{\frac{\pi}{7}}} = \frac {2\left( \cos {\frac {2\pi} {7}} + \frac{1}{2}\right)} {\frac{1}{2} \left( \cos{\frac{3\pi}{7}} + \cos{\frac{\pi}{7}}\right)} = \boxed{4} \end{aligned}$