(a) To express \(5 \sin x - 3 \cos x\) in the form \(R \sin(x - \alpha)\), 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, \(a = 5\) and \(b = -3\).

Calculate \(R\):

\(R = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{-3}{5}\)

\(\alpha = \arctan\left(\frac{-3}{5}\right)\)

\(\alpha \approx 0.54\) radians (to 2 decimal places).

(b) The maximum value of \(R \sin(x - \alpha)\) is \(R\) and the minimum value is \(-R\).

Thus, the greatest value of \((5 \sin x - 3 \cos x)^2\) is:

\(R^2 = (\sqrt{34})^2 = 34\)

The least value is:

\(0\)