To find the point where the tangent is parallel to the \(x\)-axis, we need to find where the derivative \(\frac{dy}{dx} = 0\).

Using the product rule, the derivative of \(y = e^{-ax} \tan x\) is:

\(\frac{dy}{dx} = e^{-ax} \cdot \frac{d}{dx}(\tan x) + \tan x \cdot \frac{d}{dx}(e^{-ax})\).

\(\frac{dy}{dx} = e^{-ax} \sec^2 x - ae^{-ax} \tan x\).

Setting \(\frac{dy}{dx} = 0\):

\(e^{-ax} \sec^2 x - ae^{-ax} \tan x = 0\).

\(\sec^2 x = a \tan x\).

Using \(\sec^2 x = 1 + \tan^2 x\), we have:

\(1 + \tan^2 x = a \tan x\).

Rearranging gives the quadratic equation:

\(\tan^2 x - a \tan x + 1 = 0\).

For the quadratic to have only one root, the discriminant must be zero:

\(b^2 - 4ac = 0\).

\((-a)^2 - 4 \cdot 1 \cdot 1 = 0\).

\(a^2 - 4 = 0\).

\(a^2 = 4\).

\(a = 2\) (since \(a\) is positive).

Substitute \(a = 2\) back into the quadratic equation:

\(\tan^2 x - 2 \tan x + 1 = 0\).

\((\tan x - 1)^2 = 0\).

\(\tan x = 1\).

\(x = \frac{\pi}{4}\) (within the interval \(0 < x < \frac{1}{2}\pi\)).

Thus, \(a = 2\) and \(x = \frac{1}{4}\pi\).