To find the volume of the solid formed when the shaded region is rotated about the y-axis, we use the formula for the volume of revolution:

\(V = \pi \int_{a}^{b} x^2 \, dy\)

Given \(x = \frac{12}{y^2} - 2\), we have:

\(x^2 = \left( \frac{12}{y^2} - 2 \right)^2\)

Expanding, \(x^2 = \frac{144}{y^4} - \frac{48}{y^2} + 4\).

The volume is:

\(V = \pi \int_{1}^{2} \left( \frac{144}{y^4} - \frac{48}{y^2} + 4 \right) \, dy\)

Integrating term by term:

\(\int \frac{144}{y^4} \, dy = -\frac{144}{3y^3} = -\frac{48}{y^3}\)

\(\int \frac{48}{y^2} \, dy = -\frac{48}{y}\)

\(\int 4 \, dy = 4y\)

Thus,

\(V = \pi \left[ -\frac{48}{y^3} - \frac{48}{y} + 4y \right]_{1}^{2}\)

Evaluating from 1 to 2:

\(V = \pi \left( \left( -\frac{48}{8} - \frac{48}{2} + 8 \right) - \left( -48 - 48 + 4 \right) \right)\)

\(V = \pi \left( -6 - 24 + 8 + 92 \right)\)

\(V = \pi \times 22\)

Therefore, the volume is \(22\pi\).

(i) The curve is given by \(y = \sqrt{9 - 2x^2}\). The derivative is \(\frac{dy}{dx} = \frac{1}{2\sqrt{9 - 2x^2}} \times (-4x)\).

At \(P(2, 1)\), \(x = 2\), so \(m = -4\). The gradient of the normal is \(\frac{1}{4}\).

The equation of the normal at \(P\) is \(y - 1 = \frac{1}{4}(x - 2)\).

Solving for \(A\) and \(B\):

\(A(-2, 0)\) and \(B(0, \frac{1}{2})\).

The midpoint of \(AP\) is \((0, \frac{1}{2})\), which is \(B\).

(ii) The volume of the solid of revolution is given by:

\(\int x^2 \, dy = \int \left( \frac{9}{2} - \frac{y^2}{2} \right) \, dy\).

\(= \frac{9y}{2} - \frac{y^3}{6}\).

Using limits from 1 to 3:

Volume = \(\left[ \frac{9y}{2} - \frac{y^3}{6} \right]_1^3 = \frac{44}{3} \pi\).

(i) To show that \(PQ\) is a normal to the curve, we first find the derivative of the curve \(y = (1 + 4x)^{\frac{1}{2}}\).

The derivative is \(\frac{dy}{dx} = \frac{1}{2}(1 + 4x)^{-\frac{1}{2}} \times 4 = 2(1 + 4x)^{-\frac{1}{2}}\).

At \(x = 6\), \(\frac{dy}{dx} = \frac{2}{5}\).

The gradient of the normal at \(P\) is \(-\frac{1}{2}\).

The gradient of \(PQ\) is \(-\frac{5}{2}\), hence \(PQ\) is a normal, or \(m_1 m_2 = -1\).

(ii) To find the volume of revolution, we calculate the volume for the curve and the line separately.

Volume for the curve: \(V = \pi \int (1 + 4x) \, dx\).

Integrating \(y^2 = (1 + 4x)\), we get \(\pi \left[ x + 2x^2 \right]\) from 0 to 6.

\(= \pi [6 + 72 - 0] = 78\pi\).

Volume for the line: \(V = \frac{1}{3} \pi \times 5^2 \times 2 = \frac{50}{3} \pi\).

Total volume: \(78\pi + \frac{50}{3}\pi = 94\frac{2}{3}\pi\) or \(\frac{284\pi}{3}\).

(i) To find the volume of the solid of revolution, use the formula:

\(V = \pi \int_{1}^{2} y^2 \, dx\)

Substitute \(y = \frac{4}{2x-1}\):

\(V = \pi \int_{1}^{2} \left( \frac{4}{2x-1} \right)^2 \, dx = \pi \int_{1}^{2} \frac{16}{(2x-1)^2} \, dx\)

Integrate:

\(\int \frac{16}{(2x-1)^2} \, dx = -\frac{16}{2x-1}\)

Evaluate from 1 to 2:

\(V = \pi \left[ -\frac{16}{2x-1} \right]_{1}^{2} = \pi \left( -\frac{16}{3} + 16 \right) = \frac{16\pi}{3}\)

(ii) The slope of the tangent to the curve is given by the derivative:

\(\frac{dy}{dx} = \frac{-4}{(2x-1)^2}\)

The slope of the normal is \(m = -\frac{1}{2} \times (-2) = 2\).

Equating \(\frac{dy}{dx} = -2\):

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

Solve for \(x\):

\(x = \frac{3}{2} \text{ or } x = -\frac{1}{2}\)

For \(x = \frac{3}{2}\), \(y = 2\), so \(2y = x + c\) gives \(4 = \frac{3}{2} + c\), \(c = \frac{5}{2}\).

For \(x = -\frac{1}{2}\), \(y = -2\), so \(2y = x + c\) gives \(-4 = -\frac{1}{2} + c\), \(c = -\frac{7}{2}\).

To find the volume of the solid of revolution when the region is rotated about the y-axis, we use the formula:

\(V = \pi \int x^2 \, dy\)

Given \(y = x^2 + 1\), we have \(x^2 = y - 1\).

Thus, the integral becomes:

\(V = \pi \int (y - 1) \, dy\)

We integrate with respect to \(y\) from 1 to 5:

\(V = \pi \left[ \frac{1}{2}y^2 - y \right]_1^5\)

Calculating the definite integral:

\(V = \pi \left( \frac{1}{2}(5)^2 - 5 - \left( \frac{1}{2}(1)^2 - 1 \right) \right)\)

\(V = \pi \left( \frac{25}{2} - 5 - \frac{1}{2} + 1 \right)\)

\(V = \pi \left( 12.5 - 5 - 0.5 + 1 \right)\)

\(V = \pi \times 8\)

\(V = 8\pi\)

Thus, the volume is \(8\pi\) or approximately 25.1.