(a) Differentiate the curve equation: \(\frac{dy}{dx} = 3x^2 - 4bx + b^2\).

Set \(\frac{dy}{dx} = 0\) to find critical points: \(3x^2 - 4bx + b^2 = 0\).

Factorize: \((3x - b)(x - b) = 0\), giving \(x = \frac{b}{3}\) or \(x = b\).

Since \(x = a\) at maximum, \(a = \frac{b}{3}\), hence \(b = 3a\).

(b) Find the area under the curve from \(0\) to \(a\):

\(\int (x^3 - 6ax^2 + 9a^2x) \, dx = \left[ \frac{x^4}{4} - 2ax^3 + \frac{9a^2x^2}{2} \right]_0^a\).

Evaluate: \(\frac{a^4}{4} - 2a^4 + \frac{9a^4}{2} = \frac{11a^4}{4}\).

Area under line \(OA\) is \(\frac{1}{2} a \times 4a^3 = 2a^4\).

Shaded area: \(\frac{11a^4}{4} - 2a^4 = \frac{3a^4}{4}\).

Thus, \(k = \frac{3}{4}\).