(a) Comparing coefficients gives \(R\cos\alpha=\sqrt6\), \(R\sin\alpha=3\). Thus \[\boxed{R=\sqrt{15}},\qquad\boxed{\alpha=\tan^{-1}(3/\sqrt6)=50.77^\circ}.\]

(b) Keep the unrounded angle. Let \(\beta=\cos^{-1}(2.5/\sqrt{15})\approx49.7970^\circ\). Then \(x/3-\alpha=\pm\beta\). Both lie in the required range after adding \(\alpha\), giving \[\boxed{x\approx2.9^\circ,301.7^\circ}.\]

(a) \(R\cos\alpha=\sqrt2\), \(R\sin\alpha=\sqrt5\), so \[\boxed{R=\sqrt7},\qquad\boxed{\alpha=\tan^{-1}\sqrt{5/2}\approx57.688^\circ}.\]

(b) \(\cos(2\theta+\alpha)=1/\sqrt7\). Let \(\beta=\cos^{-1}(1/\sqrt7)\approx67.7923^\circ\). In the relevant interval for \(2\theta+\alpha\), its two possible values are \(\beta\) and \(360^\circ-\beta\). Using unrounded values gives \[\boxed{\theta\approx5.1^\circ,117.3^\circ}.\]