(a) To find the stationary point, we need to differentiate \(y = x \sqrt{\sin x}\) and set the derivative to zero. Using the product rule:

\(\frac{dy}{dx} = \sqrt{\sin x} + x \cdot \frac{1}{2\sqrt{\sin x}} \cdot \cos x\)

\(= \sqrt{\sin x} + \frac{x \cos x}{2\sqrt{\sin x}}\)

Setting \(\frac{dy}{dx} = 0\):

\(\sqrt{\sin x} + \frac{x \cos x}{2\sqrt{\sin x}} = 0\)

\(2\sin x + x \cos x = 0\)

\(\tan x = -\frac{1}{2}x\)

(b) Calculate \(\tan 2\) and \(\tan 2.5\):

\(\tan 2 \approx -2.18\) and \(\tan 2.5 \approx -0.747\)

Since \(-2.18 < -1 < -0.747\), \(a\) lies between 2 and 2.5.

(c) The iterative formula is \(x_{n+1} = \pi - \arctan\left(\frac{1}{2}x_n\right)\). If it converges, it converges to \(a\) where \(\tan a = -\frac{1}{2}a\).

(d) Using the iterative formula:

Start with \(x_0 = 2.25\):

\(x_1 = \pi - \arctan\left(\frac{1}{2} \times 2.25\right) \approx 2.2974\)

\(x_2 = \pi - \arctan\left(\frac{1}{2} \times 2.2974\right) \approx 2.2871\)

\(x_3 = \pi - \arctan\left(\frac{1}{2} \times 2.2871\right) \approx 2.2893\)

\(x_4 = \pi - \arctan\left(\frac{1}{2} \times 2.2893\right) \approx 2.2888\)

Thus, \(a \approx 2.29\) to 2 decimal places.