(i) To find the coordinates of \(C\), we first find the derivative of the curve \(y = \frac{8}{\sqrt{x}} - x\). The derivative is \(\frac{dy}{dx} = -4x^{-\frac{3}{2}} - 1\). Substituting \(x = 4\) into the derivative gives \(\frac{dy}{dx} = -\frac{3}{2}\). The equation of the tangent at \(B(4, 0)\) is \(y - 0 = -\frac{3}{2}(x - 4)\). Solving for \(y\) when \(x = 1\), we find \(C(1, \frac{4}{2})\).

(ii) To find the area of the shaded region, we calculate the area under the curve from \(x = 1\) to \(x = 4\). The integral is \(\int \left( \frac{8}{\sqrt{x}} - x \right) \, dx\), which evaluates to \(8x^{\frac{1}{2}} - \frac{1}{2}x^2\). Using limits 1 to 4, the area under the curve is \(8\frac{1}{2}\). The area under the tangent is \(\frac{1}{2} \times \frac{4}{2} \times 3 = 6\frac{3}{4}\). The shaded area is \(\frac{13}{4}\).

(i) To find the equation of \(BC\), we first find the derivative of \(y = \sqrt{1 + 4x}\). The derivative is \(\frac{dy}{dx} = \frac{1}{2}(1 + 4x)^{-\frac{1}{2}} \times 4\).

At \(B(0, 1)\), the gradient of the tangent is \(2\). Therefore, the gradient of the normal is \(-\frac{1}{2}\) (since \(m_1 m_2 = -1\)).

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

(ii) To find the area of the shaded region, we first find the x-coordinate of \(A\) by setting \(y = 0\) in \(y = \sqrt{1 + 4x}\), giving \(x = -\frac{1}{4}\).

The integral of \(\sqrt{1 + 4x}\) is \(\int \sqrt{1 + 4x} \, dx = \frac{(1 + 4x)^{\frac{3}{2}}}{\frac{3}{2}} \times \frac{1}{4}\).

Evaluating from \(-\frac{1}{4}\) to \(0\), the area under the curve is \(\frac{1}{6}\).

The area of \(\triangle BOC\) is \(\frac{1}{2} \times 2 \times 1 = 1\).

Thus, the shaded area is \(1 + \frac{1}{6} = \frac{7}{6}\).

(i) Differentiate \(y = (x - 2)^4\) to find the gradient of the tangent: \(\frac{dy}{dx} = 4(x - 2)^3\). At \(x = 1\), the gradient is \(-4\). The equation of the tangent is \(y - 1 = -4(x - 1)\), which gives \(B \left( \frac{5}{4}, 0 \right)\). The gradient of the normal is \(\frac{1}{4}\), so the equation of the normal is \(y - 1 = \frac{1}{4}(x - 1)\), which gives \(C \left( 0, \frac{3}{4} \right)\).

(ii) Use the distance formula: \(AC^2 = 1^2 + \left( \frac{1}{4} \right)^2\), so \(AC = \frac{\sqrt{17}}{4}\).

(iii) Integrate \((x - 2)^4\) from 1 to 2: \(\int (x - 2)^4 \, dx = \frac{(x - 2)^5}{5}\). Evaluate from 1 to 2 to get \(\frac{1}{5}\). The area of the triangle formed by the tangent and normal is \(\frac{1}{2} \times 1 \times \frac{5}{4} = \frac{5}{8}\). The area of the shaded region is \(\frac{5}{8} - \frac{1}{5} = \frac{3}{40}\) or \(0.075\).

(i) The minimum point occurs when \(y = 0\). Solving \(x(x-2)^2 = 0\) gives \(x = 0\) or \(x = 2\). Thus, \(a = 2\).

(ii) To find \(b\), differentiate \(y = x(x-2)^2\) to get \(\frac{dy}{dx} = 3x^2 - 8x + 4\). Setting \(\frac{dy}{dx} = 0\) gives \((x-2)(3x-2) = 0\), so \(x = 2\) or \(x = \frac{2}{3}\). The maximum point is at \(x = \frac{2}{3}\), so \(b = \frac{2}{3}\).

(iii) The area of the shaded region is \(\int_0^2 x(x-2)^2 \, dx\). Expanding gives \(x^3 - 4x^2 + 4x\). Integrating, we get \(\left[ \frac{x^4}{4} - \frac{4x^3}{3} + 2x^2 \right]_0^2 = \frac{4}{3}\).

(iv) The second derivative is \(\frac{d^2y}{dx^2} = 6x - 8\). Setting \(\frac{d^2y}{dx^2} = 0\) gives \(x = \frac{4}{3}\). At this point, \(\frac{dy}{dx} = -\frac{4}{3}\), so \(m = -\frac{4}{3}\).

(i) To find the intersection points, equate \(y^2 = 2x - 1\) and \(3y = 2x - 1\).

Substitute \(y = \frac{2x - 1}{3}\) into \(y^2 = 2x - 1\):

\(\left( \frac{2x - 1}{3} \right)^2 = 2x - 1\)

\(\frac{(2x - 1)^2}{9} = 2x - 1\)

\((2x - 1)^2 = 9(2x - 1)\)

\(4x^2 - 4x + 1 = 18x - 9\)

\(4x^2 - 22x + 10 = 0\)

\(2x^2 - 11x + 5 = 0\)

Solving this quadratic equation using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(x = \frac{11 \pm \sqrt{121 - 40}}{4}\)

\(x = \frac{11 \pm \sqrt{81}}{4}\)

\(x = \frac{11 \pm 9}{4}\)

\(x = 5\) or \(x = \frac{1}{2}\)

Thus, \(a = 5\).

(ii) To find the area of the shaded region, integrate the difference between the curve and the line from \(x = \frac{1}{2}\) to \(x = 5\).

The area under the curve \(y^2 = 2x - 1\) is:

\(\int_{1/2}^{5} \sqrt{2x - 1} \, dx\)

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

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

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

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

Evaluate from \(x = \frac{1}{2}\) to \(x = 5\):

\(\left[ \frac{1}{3} (2x - 1)^{3/2} \right]_{1/2}^{5}\)

\(= \frac{1}{3} [(9)^{3/2} - (0)^{3/2}]\)

\(= \frac{1}{3} [27 - 0] = 9\)

The area under the line \(3y = 2x - 1\) is:

\(\int_{1/2}^{5} \frac{2x - 1}{3} \, dx\)

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

\(= \frac{1}{3} \left[ x^2 - x \right]_{1/2}^{5}\)

\(= \frac{1}{3} \left[ 25 - 5 - \left( \frac{1}{4} - \frac{1}{2} \right) \right]\)

\(= \frac{1}{3} \left[ 20 - \left( -\frac{1}{4} \right) \right]\)

\(= \frac{1}{3} \times 20.25 = \frac{27}{4}\)

The area of the shaded region is:

\(9 - \frac{27}{4} = \frac{9}{4}\)