(a) Multiply by \(\cos x\), which is non-zero in the original equation: \[\sqrt5+\sin x=4\cos x,\qquad4\cos x-\sin x=\sqrt5.\] Expanding \(R\cos(x+\alpha)\) gives \(R\cos\alpha=4\) and \(R\sin\alpha=1\). Thus \[\boxed{R=\sqrt{17}},\qquad\alpha=\tan^{-1}(1/4)\approx\boxed{14.04^\circ}.\]

(b) The equation is \[\sqrt{17}\cos(2x+\alpha)=\sqrt5.\] Let \(\beta=\cos^{-1}(\sqrt{5/17})\). Over \(0^\circ<x<180^\circ\), the relevant angles are \(2x+\alpha=\beta\) and \(360^\circ-\beta\). Using unrounded angles, \[\boxed{x\approx21.6^\circ,\quad144.4^\circ}.\] Neither has \(\cos2x=0\), so both satisfy the original secant equation.