To find the equation of the curve, we need to integrate \(\frac{dy}{dx} = \frac{8}{(5 - 2x)^2}\).

Let \(f(x) = \int \frac{8}{(5 - 2x)^2} \, dx\).

Using substitution, let \(u = 5 - 2x\), then \(\frac{du}{dx} = -2\) or \(dx = -\frac{1}{2} \, du\).

Substitute into the integral:

\(f(x) = \int \frac{8}{u^2} \left(-\frac{1}{2}\right) \, du = -4 \int u^{-2} \, du\).

Integrate: \(f(x) = -4 \left( \frac{u^{-1}}{-1} \right) + c = \frac{4}{u} + c\).

Substitute back \(u = 5 - 2x\):

\(f(x) = \frac{4}{5 - 2x} + c\).

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

\(7 = \frac{4}{5 - 2(2)} + c\).

\(7 = \frac{4}{1} + c\).

\(7 = 4 + c\).

\(c = 3\).

Thus, the equation of the curve is \(y = \frac{4}{5 - 2x} + 3\).

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

Integrating term by term, we have:

\(y = \int (3x^2) \, dx - \int \left(\frac{2}{x^3}\right) \, dx\)

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

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

Since the curve passes through \((-1, 3)\), substitute \(x = -1\) and \(y = 3\) to find \(c\):

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

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

\(3 = c\)

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

(a) Given \(\frac{d^2y}{dx^2} = 6x\), integrate to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \int 6x \, dx = 3x^2 + c\).

Since the curve has a stationary point at \((2, -10)\), \(\frac{dy}{dx} = 0\) when \(x = 2\):

\(3(2)^2 + c = 0\)

\(12 + c = 0\)

\(c = -12\)

Thus, \(\frac{dy}{dx} = 3x^2 - 12\).

(b) Integrate \(\frac{dy}{dx} = 3x^2 - 12\) to find \(y\):

\(y = \int (3x^2 - 12) \, dx = x^3 - 12x + k\).

Substitute \((2, -10)\) into the equation:

\(-10 = 2^3 - 12 \times 2 + k\)

\(-10 = 8 - 24 + k\)

\(-10 = -16 + k\)

\(k = 6\)

Thus, \(y = x^3 - 12x + 6\).

(c) Set \(\frac{dy}{dx} = 0\) to find other stationary points:

\(3x^2 - 12 = 0\)

\(3x^2 = 12\)

\(x^2 = 4\)

\(x = \pm 2\)

Since \(x = 2\) is already given, consider \(x = -2\):

\(y = (-2)^3 - 12(-2) + 6 = -8 + 24 + 6 = 22\)

So, the point is \((-2, 22)\).

Check the nature using \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = 6x\)

At \(x = -2\), \(\frac{d^2y}{dx^2} = 6(-2) = -12 < 0\), hence a maximum.

(d) Find the equation of the tangent at the y-intercept \((0, y)\):

\(y = 0^3 - 12 \times 0 + 6 = 6\)

The slope \(\frac{dy}{dx}\) at \(x = 0\) is:

\(\frac{dy}{dx} = 3(0)^2 - 12 = -12\)

Equation of the tangent: \(y - 6 = -12(x - 0)\)

\(y = -12x + 6\)

To find \(f(x)\), we need to integrate \(f'(x) = 3x^2 - 7\).

Integrating, we get:

\(f(x) = \int (3x^2 - 7) \, dx = x^3 - 7x + c\)

We know \(f(3) = 5\), so substitute \(x = 3\) into the equation:

\(5 = 3^3 - 7 \times 3 + c\)

\(5 = 27 - 21 + c\)

\(5 = 6 + c\)

\(c = -1\)

Thus, \(f(x) = x^3 - 7x - 1\).

To find the equation of the curve, we need to integrate \(\frac{dy}{dx} = (2x + 1)^{\frac{1}{2}}\).

Integrate: \(y = \int (2x + 1)^{\frac{1}{2}} \, dx\).

Let \(u = 2x + 1\), then \(du = 2 \, dx\) or \(dx = \frac{1}{2} \, du\).

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

Integrate: \(\frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}} = \frac{1}{3} (2x + 1)^{\frac{3}{2}} + c\).

Thus, \(y = \frac{(2x + 1)^{\frac{3}{2}}}{3} + c\).

Use the point \((4, 7)\) to find \(c\):

\(7 = \frac{(2 \cdot 4 + 1)^{\frac{3}{2}}}{3} + c\).

\(7 = \frac{9^{\frac{3}{2}}}{3} + c\).

\(7 = \frac{27}{3} + c\).

\(7 = 9 + c\).

\(c = -2\).

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