(a) To find the stationary point, we need to differentiate \(y = e^{2x}(\sin x + 3 \cos x)\) and set the derivative to zero.

Using the product rule, the derivative is:

\(y' = e^{2x}(2(\sin x + 3 \cos x)) + e^{2x}(\cos x - 3 \sin x)\)

Simplifying, we get:

\(y' = e^{2x}(2 \sin x + 6 \cos x + \cos x - 3 \sin x)\)

\(y' = e^{2x}(-\sin x + 7 \cos x)\)

Setting \(y' = 0\), we have:

\(-\sin x + 7 \cos x = 0\)

\(\tan x = 7\)

Solving for \(x\) in the interval \(0 \leq x \leq \pi\), we find:

\(x = 1.43\) (to 2 decimal places)

(b) To determine the nature of the stationary point, we check the sign of \(y'\) around \(x = 1.43\).

For \(x = 1.42\), \(y' = 0.06e^{2.84} > 0\)

For \(x = 1.44\), \(y' = -0.07e^{2.88} < 0\)

Since \(y'\) changes from positive to negative, the stationary point at \(x = 1.43\) is a maximum point.