(i) The gradient of the normal line \(3y + x = 17\) is \(-\frac{1}{3}\). Since the line is normal to the curve, the product of the gradients of the tangent and the normal is \(-1\). Therefore, the gradient of the tangent \(\frac{dy}{dx} = 3\).

Setting \(5 - \frac{8}{x^2} = 3\), we solve for \(x\):

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

\(\frac{8}{x^2} = 2\)

\(x^2 = 4\)

\(x = 2\) (since \(x\) is positive)

Substitute \(x = 2\) into the normal line equation to find \(y\):

\(3y + 2 = 17\)

\(3y = 15\)

\(y = 5\)

Thus, the coordinates of \(P\) are \((2, 5)\).

(ii) To find the equation of the curve, integrate \(\frac{dy}{dx} = 5 - \frac{8}{x^2}\):

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

\(y = 5x + 8x^{-1} + c\)

Using the point \((2, 5)\) to find \(c\):

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

\(5 = 10 + 4 + c\)

\(c = -9\)

Therefore, the equation of the curve is \(y = 5x + \frac{8}{x} - 9\).

(i) To find the stationary point, set the derivative equal to zero:

\(f'(x) = 2x - 6 = 0\)

Solving for \(x\), we get:

\(2x = 6\)

\(x = 3\)

Thus, the value of \(x\) for which \(f(x)\) has a stationary value is 3.

(ii) To find \(f(x)\), integrate \(f'(x)\):

\(f(x) = \int (2x - 6) \, dx\)

\(f(x) = x^2 - 6x + c\)

Given \(f(x) \geq -4\), substitute \(x = 3\) to find \(c\):

\(f(3) = 3^2 - 6(3) + c = -4\)

\(9 - 18 + c = -4\)

\(c = 5\)

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

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = \frac{3}{(1 + 2x)^2}\).

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

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

Substitute back \(u = 1 + 2x\):

\(y = -\frac{3}{2(1 + 2x)} + c\).

Using the point \((1, \frac{1}{2})\):

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

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

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

\(c = 1\).

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

(ii) We need \(\frac{3}{(1 + 2x)^2} < \frac{1}{3}\).

\(9 < (1 + 2x)^2\).

\((1 + 2x)^2 > 9\).

\(1 + 2x > 3\) or \(1 + 2x < -3\).

\(2x > 2\) or \(2x < -4\).

\(x > 1\) or \(x < -2\).

Thus, the set of values of \(x\) is \(x > 1, x < -2\).

(i) At \(x = 2\), the tangent has gradient \(3\). Therefore, the normal has gradient \(-\frac{1}{3}\) since \(m_1 m_2 = -1\).

The equation of the normal is \(y - 11 = -\frac{1}{3}(x - 2)\).

(ii) To find the equation of the curve, integrate \(\frac{dy}{dx} = \frac{6}{\sqrt{3x - 2}}\).

Integrate: \(y = 6 \int \frac{1}{\sqrt{3x - 2}} \, dx\).

This gives \(y = 4\sqrt{3x - 2} + c\).

Using the point \((2, 11)\) to find \(c\):

\(11 = 4\sqrt{3(2) - 2} + c\)

\(11 = 4\sqrt{4} + c\)

\(11 = 8 + c\)

\(c = 3\)

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

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = 3x^{\frac{1}{2}} - 6\).

\(y = \int (3x^{\frac{1}{2}} - 6) \, dx\)

\(y = \int 3x^{\frac{1}{2}} \, dx - \int 6 \, dx\)

\(y = 2x^{\frac{3}{2}} - 6x + c\)

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

\(2 = 2(9)^{\frac{3}{2}} - 6(9) + c\)

\(2 = 54 - 54 + c\)

\(c = 2\)

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

(ii) To find the stationary point, set \(\frac{dy}{dx} = 0\):

\(3x^{\frac{1}{2}} - 6 = 0\)

\(3x^{\frac{1}{2}} = 6\)

\(x^{\frac{1}{2}} = 2\)

\(x = 4\)

To determine the nature, find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = \frac{3}{2}x^{-\frac{1}{2}}\)

At \(x = 4\), \(\frac{d^2y}{dx^2} = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4}\)

Since \(\frac{d^2y}{dx^2} > 0\), the stationary point is a minimum.