(a) To find \(\frac{dy}{dx}\), use the product rule on \(y = e^{-4x} \tan x\):

\(\frac{dy}{dx} = \frac{d}{dx}(e^{-4x}) \tan x + e^{-4x} \frac{d}{dx}(\tan x)\).

\(\frac{d}{dx}(e^{-4x}) = -4e^{-4x}\) and \(\frac{d}{dx}(\tan x) = \sec^2 x\).

Thus, \(\frac{dy}{dx} = -4e^{-4x} \tan x + e^{-4x} \sec^2 x\).

Factor out \(e^{-4x} \sec^2 x\):

\(\frac{dy}{dx} = e^{-4x} \sec^2 x (-4 \sin x \cos x + 1)\).

Using \(\sin 2x = 2 \sin x \cos x\), rewrite as:

\(\frac{dy}{dx} = e^{-4x} \sec^2 x (1 - 2 \sin 2x)\).

Thus, \(a = 1\) and \(b = -2\).

(b) Set \(\frac{dy}{dx} = 0\):

\(\sec^2 x (1 - 2 \sin 2x) e^{-4x} = 0\).

This implies \(1 - 2 \sin 2x = 0\), so \(\sin 2x = \frac{1}{2}\).

\(2x = \frac{\pi}{6}\) or \(2x = \frac{5\pi}{6}\).

Thus, \(x = \frac{\pi}{12}\) or \(x = \frac{5\pi}{12}\).