(i) We want to express \(\sqrt{5} \cos x + 2 \sin x\) in the form \(R \cos(x - \alpha)\).

Using the identity:

\(\begin{aligned} a \cos \theta + b \sin \theta &\equiv R \cos(\theta - \alpha), \\ R &= \sqrt{a^2 + b^2}, \\ \tan \alpha &= \frac{b}{a}, \quad 0^\circ < \alpha < 90^\circ. \end{aligned}\)

Here, \(a = \sqrt{5}\) and \(b = 2\).

Calculate \(R\):

\(R = \sqrt{(\sqrt{5})^2 + 2^2} = \sqrt{5 + 4} = \sqrt{9} = 3.\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{2}{\sqrt{5}},\)

\(\alpha = \tan^{-1}\left(\frac{2}{\sqrt{5}}\right) \approx 41.81^\circ.\)

(ii) Solve \(\sqrt{5} \cos \frac{1}{2}x + 2 \sin \frac{1}{2}x = 1.2\).

We have \(R \cos\left(\frac{1}{2}x - 41.81^\circ\right) = 1.2\).

\(3 \cos\left(\frac{1}{2}x - 41.81^\circ\right) = 1.2\)

\(\cos\left(\frac{1}{2}x - 41.81^\circ\right) = 0.4\)

\(\frac{1}{2}x - 41.81^\circ = \cos^{-1}(0.4) \approx 66.42^\circ\)

\(\frac{1}{2}x = 66.42^\circ + 41.81^\circ = 108.23^\circ\)

\(x = 2 \times 108.23^\circ = 216.46^\circ\)

Rounding to one decimal place, \(x = 216.5^\circ\).