(i) To express \(8 \cos \theta + 15 \sin \theta\) in the form \(R \cos(\theta - \alpha)\), we use the identities:

\(\begin{aligned} a \cos \theta + b \sin \theta &\equiv R \cos(\theta - \alpha), \\ R &= \sqrt{a^2 + b^2}, \\ \tan \alpha &= \frac{b}{a}. \end{aligned}\)

Here, \(a = 8\) and \(b = 15\).

Calculate \(R\):

\(R = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17.\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{15}{8},\)

\(\alpha = \tan^{-1}\left(\frac{15}{8}\right) \approx 61.93^\circ.\)

(ii) Solve \(8 \cos \theta + 15 \sin \theta = 12\):

Using \(R \cos(\theta - \alpha) = 12\):

\(17 \cos(\theta - 61.93^\circ) = 12,\)

\(\cos(\theta - 61.93^\circ) = \frac{12}{17}.\)

Calculate \(\theta - 61.93^\circ\):

\(\theta - 61.93^\circ = \cos^{-1}\left(\frac{12}{17}\right) \approx 45.099^\circ.\)

Thus, \(\theta = 45.099^\circ + 61.93^\circ = 107.0^\circ\).

Also, \(\theta - 61.93^\circ = 360^\circ - 45.099^\circ = 314.901^\circ\).

Thus, \(\theta = 314.901^\circ + 61.93^\circ = 376.831^\circ\), which is outside the interval \(0^\circ < \theta < 360^\circ\).

Therefore, the solutions are \(\theta = 107.0^\circ\) and \(\theta = 16.8^\circ\).