(i) To express \(4 \cos \theta + 3 \sin \theta\) in the form \(R \cos(\theta - \alpha)\), we use the identity:

\(R \cos(\theta - \alpha) = R \cos \alpha \cos \theta + R \sin \alpha \sin \theta\).

Matching coefficients, we have:

\(R \cos \alpha = 4\) and \(R \sin \alpha = 3\).

Solving for \(R\), we get:

\(R = \sqrt{4^2 + 3^2} = 5\).

Then, \(\cos \alpha = \frac{4}{5}\) and \(\sin \alpha = \frac{3}{5}\).

\(\alpha = \arctan\left(\frac{3}{4}\right) \approx 0.6435\).

(ii) (a) Solve \(4 \cos \theta + 3 \sin \theta = 2\):

Using \(R \cos(\theta - \alpha) = 2\), we have:

\(5 \cos(\theta - 0.6435) = 2\).

\(\cos(\theta - 0.6435) = \frac{2}{5}\).

\(\theta - 0.6435 = \arccos\left(\frac{2}{5}\right)\).

\(\theta = 0.6435 + \arccos\left(\frac{2}{5}\right) \approx 1.80\).

Also, \(\theta = 0.6435 + 2\pi - \arccos\left(\frac{2}{5}\right) \approx 5.77\).

(b) Find \(\int \frac{50}{(4 \cos \theta + 3 \sin \theta)^2} \, d\theta\):

Express the integrand as \(k \sec^2(\theta - 0.6435)\) where \(k = 2\).

\(\int 2 \sec^2(\theta - 0.6435) \, d\theta = 2 \tan(\theta - 0.6435) + C\).