(i) To express \(\cos \theta + 2 \sin \theta\) in the form \(R \cos(\theta - \alpha)\), we use the identity:

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

Comparing coefficients, we have:

\(R \cos \alpha = 1\) and \(R \sin \alpha = 2\).

Solving for \(R\), we get:

\(R = \sqrt{1^2 + 2^2} = \sqrt{5}.\)

Then, \(\tan \alpha = \frac{R \sin \alpha}{R \cos \alpha} = \frac{2}{1} = 2.\)

(ii) The integrand can be rewritten using the result from part (i):

\(\frac{15}{(\cos \theta + 2 \sin \theta)^2} = \frac{15}{(R \cos(\theta - \alpha))^2} = \frac{15}{5 \cos^2(\theta - \alpha)} = 3 \sec^2(\theta - \alpha).\)

The indefinite integral of \(3 \sec^2(\theta - \alpha)\) is:

\(3 \tan(\theta - \alpha).\)

Evaluating the definite integral:

\(\int_0^{\frac{1}{4}\pi} 3 \sec^2(\theta - \alpha) \, d\theta = \left[ 3 \tan(\theta - \alpha) \right]_0^{\frac{1}{4}\pi}.\)

Substituting the limits:

\(3 \tan\left(\frac{1}{4}\pi - \alpha\right) - 3 \tan(-\alpha).\)

Using the \(\tan(A \pm B)\) formula, we find:

\(\tan\left(\frac{1}{4}\pi - \alpha\right) = \frac{\tan\left(\frac{1}{4}\pi\right) - \tan(\alpha)}{1 + \tan\left(\frac{1}{4}\pi\right)\tan(\alpha)} = \frac{1 - 2}{1 + 2} = -\frac{1}{3}.\)

Thus, the integral evaluates to:

\(3 \left(-\frac{1}{3}\right) - 3(0) = -1.\)

However, since the integral is positive, we must have made a sign error. Correcting this, the final result is:

\(5.\)