**(a)** To express \(\sqrt{7} \sin x + 2 \cos x\) in the form \(R \sin(x + \alpha)\), we use the identity:

\(a \sin \theta + b \cos \theta \equiv R \sin(\theta + \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = \sqrt{7}\) and \(b = 2\).

Calculate \(R\):

\(R = \sqrt{(\sqrt{7})^2 + 2^2} = \sqrt{7 + 4} = \sqrt{11}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{2}{\sqrt{7}}\)

\(\alpha = \tan^{-1}\left(\frac{2}{\sqrt{7}}\right) \approx 37.09^\circ\)

**(b)** Solve \(\sqrt{7} \sin 2\theta + 2 \cos 2\theta = 1\).

Using the result from part (a), express the equation as:

\(\sqrt{11} \sin(2\theta + 37.09^\circ) = 1\)

\(\sin(2\theta + 37.09^\circ) = \frac{1}{\sqrt{11}}\)

Find the angle:

\(2\theta + 37.09^\circ = \sin^{-1}\left(\frac{1}{\sqrt{11}}\right) \approx 17.5484^\circ\)

Thus, \(2\theta = 17.5484^\circ - 37.09^\circ\) or \(2\theta = 180^\circ - 17.5484^\circ - 37.09^\circ\).

Calculate \(\theta\):

\(\theta = \frac{17.5484^\circ - 37.09^\circ}{2} \approx 62.7^\circ\)

\(\theta = \frac{180^\circ - 17.5484^\circ - 37.09^\circ}{2} \approx 170.2^\circ\)