(a) Expanding, \[R\cos(x+\alpha)=R\cos\alpha\cos x-R\sin\alpha\sin x.\] Therefore \[R\cos\alpha=4,\quad R\sin\alpha=1,\quad\boxed{R=\sqrt{17}},\quad\alpha=\tan^{-1}(1/4)\approx\boxed{14.04^\circ}.\]

(b) \[\sqrt{17}\cos(2x+\alpha)=3,\qquad\cos(2x+\alpha)=3/\sqrt{17}.\] Let \(\beta=\cos^{-1}(3/\sqrt{17})\approx43.3139^\circ\). The angle \(2x+\alpha\) lies between \(\alpha\) and \(360^\circ+\alpha\), so the relevant values are \(\beta\) and \(360^\circ-\beta\). Using unrounded angles, \[\boxed{x\approx14.6^\circ,\quad151.3^\circ}.\]