(a) At the stationary point, \(\frac{dy}{dx} = 0\). Therefore,

\(6(3x-5)^3 - kx^2 = 0\)

Substitute \(x = 2\):

\(6(3(2)-5)^3 - k(2)^2 = 0\)

\(6(1)^3 - 4k = 0\)

\(6 - 4k = 0\)

\(4k = 6\)

\(k = \frac{3}{2}\)

(b) Integrate \(\frac{dy}{dx} = 6(3x-5)^3 - \frac{3}{2}x^2\):

\(y = \frac{6}{4 \times 3}(3x-5)^4 - \frac{1}{3}kx^3 + c\)

\(y = \frac{1}{2}(3x-5)^4 - \frac{1}{2}x^3 + c\)

Use the point \((2, -3.5)\):

\(-3.5 = \frac{1}{2}(3(2)-5)^4 - \frac{1}{2}(2)^3 + c\)

\(-3.5 = \frac{1}{2}(1)^4 - 4 + c\)

\(-3.5 = \frac{1}{2} - 4 + c\)

\(-3.5 = -3.5 + c\)

\(c = 0\)

Thus, the equation of the curve is:

\(y = \frac{1}{2}(3x-5)^4 - \frac{1}{2}x^3\)