To find the equation of the curve, we need to integrate the derivative \(\frac{dy}{dx} = \frac{8}{(3x + 2)^2}\).

The integral of \(\frac{8}{(3x + 2)^2}\) is \(-\frac{8}{3(3x + 2)} + c\), where \(c\) is the constant of integration.

Given that the curve passes through the point \((2, 5\frac{2}{3})\), we substitute \(x = 2\) and \(y = 5\frac{2}{3}\) into the equation:

\(5\frac{2}{3} = -\frac{8}{3(3 \times 2 + 2)} + c\)

\(5\frac{2}{3} = -\frac{8}{12} + c\)

\(5\frac{2}{3} = -\frac{2}{3} + c\)

Solving for \(c\), we get \(c = 6\).

Thus, the equation of the curve is \(y = -\frac{8}{3(3x+2)} + 6\).

To find \(f(x)\), we need to integrate \(f'(x) = 6x^2 - \frac{8}{x^2}\).

Integrate term by term:

\(\int 6x^2 \, dx = 2x^3 + C_1\)

\(\int -\frac{8}{x^2} \, dx = 8x^{-1} + C_2\)

Thus, \(f(x) = 2x^3 + \frac{8}{x} + C\).

Given that the curve passes through \((2, 7)\), substitute \(x = 2\) and \(f(x) = 7\):

\(7 = 2(2)^3 + \frac{8}{2} + C\)

\(7 = 16 + 4 + C\)

\(7 = 20 + C\)

\(C = -13\)

Therefore, \(f(x) = 2x^3 + \frac{8}{x} - 13\).

(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:

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

To find the equation of the curve, we need to integrate the derivative \(\frac{dy}{dx} = \frac{3}{x^4} + 32x^3\).

Integrating \(\frac{3}{x^4}\) gives \(-\frac{1}{x^3}\).

Integrating \(32x^3\) gives \(8x^4\).

Thus, the general solution is \(y = -\frac{1}{x^3} + 8x^4 + c\).

We use the point \(\left( \frac{1}{2}, 4 \right)\) to find \(c\).

Substitute \(x = \frac{1}{2}\) and \(y = 4\) into the equation:

\(4 = -\frac{1}{(\frac{1}{2})^3} + 8(\frac{1}{2})^4 + c\)

\(4 = -8 + \frac{1}{2} + c\)

\(4 = -8 + 0.5 + c\)

\(4 = -7.5 + c\)

\(c = 4 + 7.5 = 11.5\)

Therefore, the equation of the curve is \(y = -\frac{1}{x^3} + 8x^4 + \frac{23}{2}\).

To find the equation of the curve, we need to integrate \(\frac{dy}{dx} = \frac{6}{(3x - 2)^3}\).

Let \(u = 3x - 2\), then \(\frac{du}{dx} = 3\) or \(dx = \frac{du}{3}\).

Substitute into the integral:

\(\int \frac{6}{u^3} \cdot \frac{du}{3} = 2 \int u^{-3} \, du\).

The integral of \(u^{-3}\) is \(\frac{u^{-2}}{-2}\), so:

\(2 \left( \frac{u^{-2}}{-2} \right) = -u^{-2} + c\).

Substitute back \(u = 3x - 2\):

\(y = -(3x-2)^{-2} + c\).

Using the point \(A(1, -3)\), substitute \(x = 1\) and \(y = -3\):

\(-3 = -(3(1)-2)^{-2} + c\).

\(-3 = -1 + c\).

\(c = -2\).

Thus, the equation of the curve is:

\(y = -(3x-2)^{-2} - 2\).