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

\(\int (5 - 2x^2) \, dx = \int 5 \, dx - \int 2x^2 \, dx\)

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

We know that \((3, 5)\) is a point on the curve, so \(f(3) = 5\).

Substitute \(x = 3\) and \(f(x) = 5\) into the equation:

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

\(5 = 15 - \frac{54}{3} + c\)

\(5 = 15 - 18 + c\)

\(5 = -3 + c\)

\(c = 8\)

Thus, \(f(x) = 5x - \frac{2x^3}{3} + 8\).

To find \(f(x)\), we need to integrate \(f'(x) = x^{-\frac{3}{2}} + 1\).

The integral of \(x^{-\frac{3}{2}}\) is \(2x^{-\frac{1}{2}}\), and the integral of \(1\) is \(x\).

Thus, \(f(x) = 2x^{\frac{1}{2}} + x + c\), where \(c\) is a constant.

We use the condition \(f(4) = 5\) to find \(c\):

\(5 = 2 \times 4^{\frac{1}{2}} + 4 + c\)

\(5 = 2 \times 2 + 4 + c\)

\(5 = 4 + 4 + c\)

\(5 = 8 + c\)

\(c = 5 - 8 = -3\)

Therefore, \(f(x) = 2x^{\frac{1}{2}} + x - 3\).

To find \(f(x)\), we need to integrate \(f'(x) = \frac{1}{\sqrt{x+6}} + \frac{6}{x^2}\).

The integral of \(\frac{1}{\sqrt{x+6}}\) is \(2\sqrt{x+6}\).

The integral of \(\frac{6}{x^2}\) is \(-\frac{6}{x}\).

Thus, \(f(x) = 2\sqrt{x+6} - \frac{6}{x} + c\).

Using the condition \(f(3) = 1\), substitute \(x = 3\) into the equation:

\(2\sqrt{3+6} - \frac{6}{3} + c = 1\).

This simplifies to \(2(3) - 2 + c = 1\).

\(6 - 2 + c = 1\).

\(c = -3\).

Therefore, \(f(x) = 2\sqrt{x+6} - \frac{6}{x} - 3\).

Given \(\frac{dy}{dx} = \sqrt{2x + 5}\), we need to find \(y\) by integrating.

Integrate \(\frac{dy}{dx} = \sqrt{2x + 5} = (2x + 5)^{1/2}\).

\(y = \int (2x + 5)^{1/2} \, dx\)

Using the substitution \(u = 2x + 5\), \(du = 2 \, dx\), so \(dx = \frac{1}{2} \, du\).

\(y = \int (u)^{1/2} \cdot \frac{1}{2} \, du\)

\(y = \frac{1}{2} \int u^{1/2} \, du\)

\(y = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + c\)

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

Given the point \((2, 5)\) is on the curve, substitute \(x = 2\) and \(y = 5\):

\(5 = \frac{1}{3} (2(2) + 5)^{3/2} + c\)

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

\(5 = \frac{1}{3} (9)^{3/2} + c\)

\(5 = \frac{1}{3} \cdot 27 + c\)

\(5 = 9 + c\)

\(c = -4\)

Thus, the equation of the curve is:

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

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

Given \(\frac{dy}{dx} = \frac{6}{x^2}\), we need to find \(y\) by integrating \(\frac{dy}{dx}\).

Integrate \(\frac{dy}{dx} = \frac{6}{x^2}\):

\(y = \int \frac{6}{x^2} \, dx = \int 6x^{-2} \, dx\)

\(y = 6 \cdot \left( \frac{x^{-1}}{-1} \right) + c = -6x^{-1} + c\)

Using the point \((2, 9)\) to find \(c\):

\(9 = -6(2)^{-1} + c\)

\(9 = -3 + c\)

\(c = 12\)

Thus, the equation of the curve is \(y = -6x^{-1} + 12\).