(a) Expansion gives \(4\cos x+\sqrt3\sin x\). Comparing with \(R\cos\alpha\cos x+R\sin\alpha\sin x\), \(R\cos\alpha=4,\ R\sin\alpha=\sqrt3\). Thus \(\boxed{R=\sqrt{19},\ \alpha=\tan^{-1}(\sqrt3/4)=23.41^\circ}\).

(b) \(\cos(2\theta-\alpha)=2.5/\sqrt{19}\). Let \(\beta=\cos^{-1}(2.5/\sqrt{19})\). In \(-\alpha<2\theta-\alpha<360^\circ-\alpha\), the solutions are \(\beta,360^\circ-\beta\). Therefore \(\theta=(\alpha+\beta)/2\) or \((\alpha+360^\circ-\beta)/2\), giving \(\boxed{\theta=39.2^\circ,\ 164.2^\circ}\) (1 d.p.).