(a) To find the stationary point, we need to differentiate \(y = \frac{x^3}{e^x - 1}\) and set the derivative to zero.

Using the quotient rule, \(\frac{d}{dx} \left( \frac{x^3}{e^x - 1} \right) = \frac{(e^x - 1)(3x^2) - x^3(e^x)}{(e^x - 1)^2}\).

Setting the numerator to zero for a stationary point: \((e^x - 1)(3x^2) - x^3(e^x) = 0\).

Simplifying gives \(3x^2 e^x - 3x^2 - x^3 e^x = 0\).

Rearranging, \(x^2 e^x (3 - x) = 3x^2\).

Dividing by \(x^2\) (assuming \(x \neq 0\)), \(e^x (3 - x) = 3\).

Thus, \(3 - x = 3e^{-x}\) or \(x = 3(1 - e^{-x})\).

Therefore, \(p = 3(1 - e^{-p})\).

(b) To verify \(p\) lies between 2.5 and 3, calculate \(f(p) = 3(1 - e^{-p}) - p\).

For \(p = 2.5\), \(f(2.5) = 3(1 - e^{-2.5}) - 2.5 \approx 0.351\).

For \(p = 3\), \(f(3) = 3(1 - e^{-3}) - 3 \approx -0.950\).

Since \(f(2.5) > 0\) and \(f(3) < 0\), by the Intermediate Value Theorem, \(p\) lies between 2.5 and 3.

(c) Using the iterative formula \(p_{n+1} = 3(1 - e^{-p_n})\), start with an initial guess, say \(p_0 = 2.5\).

Calculate successive iterations:

\(p_1 = 3(1 - e^{-2.5}) \approx 2.7639\)

\(p_2 = 3(1 - e^{-2.7639}) \approx 2.8113\)

\(p_3 = 3(1 - e^{-2.8113}) \approx 2.8212\)

\(p_4 = 3(1 - e^{-2.8212}) \approx 2.8231\)

\(p_5 = 3(1 - e^{-2.8231}) \approx 2.8235\)

The iterations converge to \(p = 2.82\) to 2 decimal places.