(i) To express \(4 \sin \theta - 3 \cos \theta\) in the form \(R \sin(\theta - \alpha)\), 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, \(a = 4\) and \(b = 3\).

Calculate \(R\):

\(R = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{3}{4}\)

\(\alpha = \tan^{-1}\left(\frac{3}{4}\right) \approx 36.87^\circ\)

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

Using the form \(R \sin(\theta - \alpha)\), we have:

\(5 \sin(\theta - 36.87^\circ) = 2\)

\(\sin(\theta - 36.87^\circ) = \frac{2}{5}\)

\(\theta - 36.87^\circ = \sin^{-1}\left(\frac{2}{5}\right)\)

\(\theta - 36.87^\circ = 23.58^\circ\)

\(\theta = 23.58^\circ + 36.87^\circ = 60.45^\circ\)

For the second solution:

\(\theta = 180^\circ - 23.58^\circ + 36.87^\circ = 193.29^\circ\)

(iii) The expression \(\frac{1}{4 \sin \theta - 3 \cos \theta + 6}\) has its greatest value when \(4 \sin \theta - 3 \cos \theta\) is minimized.

The minimum value of \(4 \sin \theta - 3 \cos \theta\) is \(-5\), so:

\(4 \sin \theta - 3 \cos \theta + 6 = 1\)

Thus, the greatest value of the expression is 1.