To find \(f(x)\), we need to integrate \(f'(x) = 3x^2 + 2x - 5\).

The integral of \(3x^2\) is \(x^3\), the integral of \(2x\) is \(x^2\), and the integral of \(-5\) is \(-5x\). Therefore,

\(f(x) = x^3 + x^2 - 5x + c\), where \(c\) is the constant of integration.

We use the point \((1, 3)\) to find \(c\). Substituting \(x = 1\) and \(f(x) = 3\) into the equation:

\(3 = 1^3 + 1^2 - 5(1) + c\)

\(3 = 1 + 1 - 5 + c\)

\(3 = -3 + c\)

\(c = 6\)

Thus, \(f(x) = x^3 + x^2 - 5x + 6\).

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

First, integrate \(\frac{3}{\sqrt{x}}\):

\(\int \frac{3}{\sqrt{x}} \, dx = \int 3x^{-1/2} \, dx = 6\sqrt{x} + C_1\).

Next, integrate \(-x\):

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

Combining these, the general solution is:

\(y = 6\sqrt{x} - \frac{x^2}{2} + C\).

Given the point (4, 6), substitute \(x = 4\) and \(y = 6\) to find \(C\):

\(6 = 6\sqrt{4} - \frac{4^2}{2} + C\).

\(6 = 12 - 8 + C\).

\(C = 2\).

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

To find the equation of the curve, we need to integrate the derivative \(\frac{dy}{dx} = 2x^2 - 5\).

Integrating, we get:

\(y = \int (2x^2 - 5) \, dx = \frac{2x^3}{3} - 5x + c\)

We use the point \((3, 8)\) to find the constant \(c\).

Substitute \(x = 3\) and \(y = 8\) into the equation:

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

\(8 = \frac{54}{3} - 15 + c\)

\(8 = 18 - 15 + c\)

\(8 = 3 + c\)

\(c = 5\)

Thus, the equation of the curve is:

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

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

Integrate: \(y = \int \frac{4}{(x-3)^3} \, dx\).

Let \(u = x-3\), then \(du = dx\).

\(y = \int \frac{4}{u^3} \, du = 4 \int u^{-3} \, du\).

Integrating, we get \(y = 4 \left( \frac{u^{-2}}{-2} \right) + c = \frac{-2}{u^2} + c\).

Substitute back \(u = x-3\): \(y = \frac{-2}{(x-3)^2} + c\).

Use the point (4, 5) to find \(c\):

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

\(5 = -2 + c\).

\(c = 7\).

Thus, the equation of the curve is \(y = \frac{-2}{(x-3)^2} + 7\).

(a) To find the value of \(a\), set \(\frac{dy}{dx} = 0\) at the stationary point \(x = a\). Thus, \(6a^2 - 30a + 6a = 0\). Simplifying gives \(6a(a - 4) = 0\), so \(a = 4\).

(b) Differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2} = 12x - 30\). At \(x = 4\), \(\frac{d^2y}{dx^2} = 18\), which is positive, indicating a minimum.

(c) Integrate \(\frac{dy}{dx} = 6x^2 - 30x + 24\) to find \(y\). The integral is \(y = 2x^3 - 15x^2 + 24x + c\). Using the point \((4, -15)\), solve for \(c\): \(-15 = 2(4)^3 - 15(4)^2 + 24(4) + c\), giving \(c = 1\). Thus, \(y = 2x^3 - 15x^2 + 24x + 1\).

(d) Set \(\frac{dy}{dx} = 0\) to find other stationary points: \(6x^2 - 30x + 24 = 0\). Factor to get \(6(x-1)(x-4) = 0\), so \(x = 1\) or \(x = 4\). For \(x = 1\), \(y = 2(1)^3 - 15(1)^2 + 24(1) + 1 = 12\). Thus, the coordinates are \((1, 12)\).