(a) By the product and chain rules, \[\frac{dy}{dx}=e^{-4x}\sec^2x-4e^{-4x}\tan x.\] Factor out \(e^{-4x}\sec^2x\) and use \(\tan x/\sec^2x=\sin x\cos x=\tfrac12\sin2x\): \[\boxed{\frac{dy}{dx}=\sec^2x(1-2\sin2x)e^{-4x}}.\] Thus \(a=1\), \(b=-2\).

(b) The exponential and squared secant are positive in the interval, so \[1-2\sin2x=0,\qquad\sin2x=1/2.\] In \(0\le2x<\pi\), the solutions are \(2x=\pi/6,5\pi/6\). Hence \[\boxed{x=\frac\pi{12},\quad\frac{5\pi}{12}}.\]