(a) Start by expressing the left-hand side as a single fraction:

\(\frac{\cos 3x}{\sin x} + \frac{\sin 3x}{\cos x} = \frac{\cos 3x \cdot \cos x + \sin 3x \cdot \sin x}{\sin x \cdot \cos x}\)

Use the angle addition formula \(\cos(A - B) = \cos A \cos B + \sin A \sin B\):

\(= \frac{\cos(3x - x)}{\sin x \cos x} = \frac{\cos 2x}{\sin x \cos x}\)

Use the double angle identity \(\sin 2x = 2 \sin x \cos x\):

\(= \frac{\cos 2x}{\frac{1}{2} \sin 2x} = \frac{2 \cos 2x}{\sin 2x} = 2 \cot 2x\)

Thus, the identity is proven.

(b) Set \(\frac{\cos 3x}{\sin x} + \frac{\sin 3x}{\cos x} = 4\):

\(2 \cot 2x = 4\) \(\cot 2x = 2\)

Then, \(\tan 2x = \frac{1}{2}\).

Find \(2x\):

\(2x = \tan^{-1}\left(\frac{1}{2}\right) \approx 0.464 \text{ or } 0.464 + n\pi\)

For \(0 < x < \pi\), solve for \(x\):

\(x = \frac{0.464}{2} \approx 0.232\)

And for the next solution:

\(x = \frac{0.464 + \pi}{2} \approx 1.80\)

Thus, \(x \approx 0.232, 1.80\).