(i) To find the area of the shaded region, we calculate the integral of the difference between the curve and the line from \(x = -1\) to \(x = 0\):

\(\int_{-1}^{0} \left( (x+1)^{\frac{1}{2}} - (x+1) \right) \, dx\)

Calculate the integrals separately:

\(\int (x+1)^{\frac{1}{2}} \, dx = \frac{2}{3}(x+1)^{\frac{3}{2}}\)

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

Apply the limits:

\(\frac{2}{3}(0+1)^{\frac{3}{2}} - \left( \frac{1}{2}(0)^2 + 0 \right) - \left( \frac{2}{3}(-1+1)^{\frac{3}{2}} - \left( \frac{1}{2}(-1)^2 - 1 \right) \right)\)

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

(ii) To find the volume of the solid obtained by rotating the shaded region about the y-axis, use the method of cylindrical shells:

\(V = \pi \int_{0}^{1} \left( (y^2 - 1)^2 - (y-1)^2 \right) \, dy\)

Calculate the integrals:

\(\int (y^4 - 2y^2 + 1) \, dy = \frac{y^5}{5} - \frac{2y^3}{3} + y\)

Apply the limits:

\(\pi \left[ \frac{1}{5} - \frac{2}{3} + 1 \right] = \frac{8}{15} \pi\)

Alternatively, using the volume of a cone:

\(V = \frac{1}{3} \pi (x_1^2 - x_2^2)\)

\(= \frac{1}{5} \pi\)

(i) To find \(\frac{dy}{dx}\), use the chain rule. Let \(u = 8x - x^2\), then \(y = \sqrt{u} = u^{\frac{1}{2}}\).

\(\frac{dy}{du} = \frac{1}{2}u^{-\frac{1}{2}}\)

\(\frac{du}{dx} = 8 - 2x\)

Thus, \(\frac{dy}{dx} = \frac{1}{2}(8x-x^2)^{-\frac{1}{2}} \times (8-2x)\).

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

\((8-2x) = 0 \Rightarrow x = 4\).

Substitute \(x = 4\) into \(y = \sqrt{(8x - x^2)}\):

\(y = \sqrt{(8 \times 4 - 4^2)} = \sqrt{16} = 4\).

Thus, the stationary point is (4, 4).

(ii) To find the volume, use the formula for the volume of revolution:

\(V = \pi \int (8x - x^2) \, dx\) from \(x = 0\) to \(x = 8\).

\(V = \pi \left[ \frac{4x^2}{2} - \frac{x^3}{3} \right]_0^8\)

\(V = \pi \left[ \frac{4 \times 8^2}{2} - \frac{8^3}{3} \right]\)

\(V = \pi \left[ 128 - \frac{512}{3} \right]\)

\(V = \pi \left[ \frac{384}{3} \right]\)

\(V = \frac{256\pi}{3}\).

(i) To find the coordinates of \(B\), set \(x = 0\) in the equation \(y = \sqrt{1 + 2x}\):

\(y = \sqrt{1 + 2(0)} = \sqrt{1} = 1\)

So, \(B = (0, 1)\).

For \(C\), given \(y = 3\):

\(3 = \sqrt{1 + 2x}\)

\(9 = 1 + 2x\)

\(2x = 8\)

\(x = 4\)

So, \(C = (4, 3)\).

(ii) The derivative \(\frac{dy}{dx}\) is:

\(\frac{dy}{dx} = \frac{1}{2} \times 2(1 + 2x)^{-\frac{1}{2}} = \frac{1}{\sqrt{1 + 2x}}\)

At \(C(4, 3)\):

\(\frac{dy}{dx} = \frac{1}{\sqrt{9}} = \frac{1}{3}\)

The gradient of the normal is \(-3\).

Equation of the normal:

\(y - 3 = -3(x - 4)\)

or \(y = -3x + 15\).

(iii) To find the volume, use the formula for the volume of revolution:

\(y^2 = 1 + 2x \Rightarrow x = \frac{1}{2}(y^2 - 1)\)

Volume \(V = \pi \int_0^1 x^2 \, dy\)

\(V = \pi \int_0^1 \left( \frac{1}{2}(y^2 - 1) \right)^2 \, dy\)

\(V = \pi \times \frac{1}{4} \int_0^1 (y^4 - 2y^2 + 1) \, dy\)

\(V = \pi \times \frac{1}{4} \left[ \frac{y^5}{5} - \frac{2y^3}{3} + y \right]_0^1\)

\(V = \pi \times \frac{1}{4} \left[ \frac{1}{5} - \frac{2}{3} + 1 \right]\)

\(V = \frac{2}{15} \pi\)

(i) To find the coordinates of \(A\), set \(y = 0\) in the equation \(y = 4\sqrt{x} - x\):

\(4\sqrt{x} - x = 0\)

\(4\sqrt{x} = x\)

\(4 = \sqrt{x}\)

\(x = 16\)

Thus, \(A(16, 0)\).

To find the coordinates of \(M\), find the derivative \(\frac{dy}{dx} = 2x^{-\frac{1}{2}} - 1\) and set it to zero:

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

\(2\sqrt{x} = 1\)

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

\(x = 4\)

Substitute \(x = 4\) back into the original equation:

\(y = 4\sqrt{4} - 4 = 4\)

Thus, \(M(4, 4)\).

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

\(V = \pi \int y^2 \, dx\)

\(V = \pi \int (4\sqrt{x} - x)^2 \, dx\)

\(V = \pi \int (16x + x^2 - 8x^{\frac{3}{2}}) \, dx\)

Integrate each term:

\(\pi \left[ 8x^2 + \frac{x^3}{3} - 8\frac{x^{\frac{5}{2}}}{\frac{5}{2}} \right]\)

Evaluate from 0 to 16:

\(V = \pi \left[ 8(16)^2 + \frac{(16)^3}{3} - 8\frac{(16)^{\frac{5}{2}}}{\frac{5}{2}} \right]\)

\(V = 136.5\pi\) (or \(137\pi\))

(i) The curve \(y = (x - 2)^2\) is a parabola opening upwards with vertex at \((2, 0)\). It touches the positive \(x\)-axis at \(x = 2\).

(ii) To find the volume of the solid of revolution, use the formula for the volume of a solid of revolution about the \(x\)-axis: \(V = \pi \int (y)^2 \, dx\).

Here, \(y = (x - 2)^2\), so \(y^2 = (x - 2)^4\).

Calculate the integral: \(V = \pi \int_0^2 (x - 2)^4 \, dx\).

Expand \((x - 2)^4\) and integrate: \(\pi \left[ \frac{(x - 2)^5}{5} \right]_0^2\).

Evaluate the integral: \(\pi \left[ \frac{(2 - 2)^5}{5} - \frac{(-2)^5}{5} \right] = \pi \left[ 0 - \frac{-32}{5} \right] = \frac{32\pi}{5}\).

Thus, the volume is \(\frac{32\pi}{5}\) or \(6.4\pi\).