(i) To find the coordinates of \(A\), we need to find the maximum point of the curve \(y = -x^2 + 8x - 10\). The derivative is \(\frac{dy}{dx} = -2x + 8\). Setting \(\frac{dy}{dx} = 0\) gives \(-2x + 8 = 0\), so \(x = 4\). Substituting \(x = 4\) into the equation of the curve gives \(y = -(4)^2 + 8(4) - 10 = 6\). Thus, \(A(4, 6)\).

The gradient of line \(BA\) is 2, so the equation of line \(AB\) is \(y - 6 = 2(x - 4)\). Simplifying gives \(y = 2x - 2\).

To find \(B\), solve the system of equations: \(y = -x^2 + 8x - 10\) and \(y = 2x - 2\). Equating gives \(-x^2 + 8x - 10 = 2x - 2\), which simplifies to \(x^2 - 6x + 8 = 0\). Solving this quadratic equation gives \(x = 2\) or \(x = 4\). For \(x = 2\), \(y = 2(2) - 2 = 2\). Thus, \(B(2, 2)\).

(ii) To find the area of the shaded region, evaluate \(\int_{2}^{4} (-x^2 + 8x - 10) \, dx\). The integral is \(\int (-x^2 + 8x - 10) \, dx = -\frac{x^3}{3} + 4x^2 - 10x\).

Evaluate from 2 to 4: \(\left[-\frac{(4)^3}{3} + 4(4)^2 - 10(4)\right] - \left[-\frac{(2)^3}{3} + 4(2)^2 - 10(2)\right]\).

Calculating gives \(\left[-\frac{64}{3} + 64 - 40\right] - \left[-\frac{8}{3} + 16 - 20\right] = \left[-\frac{64}{3} + 24\right] - \left[-\frac{8}{3} - 4\right]\).

This simplifies to \(\frac{28}{3} + \frac{20}{3} = \frac{48}{3} = 16\).

Thus, the area of the shaded region is \(9\frac{1}{3}\).

(i) To find the points of intersection, set the equations equal: \(9 - x^3 = \frac{8}{x^3}\). Multiply through by \(x^3\) to get \(9x^3 - x^6 = 8\), which simplifies to \(x^6 - 9x^3 + 8 = 0\). Factor this as \((X - 1)(X - 8) = 0\) where \(X = x^3\). Thus, \(x^3 = 1\) or \(x^3 = 8\), giving \(x = 1\) or \(x = 2\). Therefore, \(a = 1\) and \(b = 2\).

(ii) The area between the curves is given by the integral \(\int_{1}^{2} \left( 9 - x^3 - \frac{8}{x^3} \right) \, dx\). Integrate to find \(\left[ 9x - \frac{x^4}{4} + \frac{4}{x^2} \right]_{1}^{2}\). Evaluating this gives \(18 - 4 + 1 - (9 - \frac{1}{4} + 4) = 2\frac{1}{4}\).

(iii) The derivatives are \(\frac{dy}{dx} = -3x^2\) for \(y = 9 - x^3\) and \(\frac{dy}{dx} = -\frac{24}{x^4}\) for \(y = \frac{8}{x^3}\). Set \(-3c^2 = -\frac{24}{c^4}\) to find \(c\). Solving gives \(c^6 = 8\), so \(c = \sqrt[6]{8}\) or approximately \(1.41 \ldots\).

First, find the points of intersection by solving the equations simultaneously:

Set \(y = (x-2)^2\) equal to \(y = 7 - 2x\):

\((x-2)^2 = 7 - 2x\)

Expand and rearrange:

\(x^2 - 4x + 4 = 7 - 2x\)

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

Factorize:

\((x-3)(x+1) = 0\)

Thus, \(x = 3\) or \(x = -1\).

Find corresponding \(y\) values:

For \(x = 3\), \(y = (3-2)^2 = 1\), so \(B(3, 1)\).

For \(x = -1\), \(y = (-1-2)^2 = 9\), so \(A(-1, 9)\).

Calculate the area under the line from \(x = -1\) to \(x = 3\):

\(\int_{-1}^{3} (7 - 2x) \, dx = [7x - x^2]_{-1}^{3}\)

\(= (21 - 9) - (-7 - 1) = 12 + 8 = 20\)

Calculate the area under the curve from \(x = -1\) to \(x = 3\):

\(\int_{-1}^{3} (x-2)^2 \, dx = \int_{-1}^{3} (x^2 - 4x + 4) \, dx\)

\(= \left[ \frac{x^3}{3} - 2x^2 + 4x \right]_{-1}^{3}\)

\(= \left( \frac{27}{3} - 18 + 12 \right) - \left( \frac{-1}{3} - 2 - 4 \right)\)

\(= (9 - 18 + 12) - (-\frac{1}{3} + 2 + 4)\)

\(= 3 - \frac{5}{3} = \frac{4}{3}\)

Find the area of the shaded region:

\(20 - \frac{4}{3} = \frac{60}{3} - \frac{4}{3} = \frac{56}{3} = 10\frac{2}{3}\)

The curve \(y = 6x - x^2\) intersects the line \(y = 5\) when:

\(6x - x^2 = 5\)

\(x^2 - 6x + 5 = 0\)

Factoring gives:

\((x - 1)(x - 5) = 0\)

Thus, \(x = 1\) and \(x = 5\).

The area under the curve from \(x = 1\) to \(x = 5\) is found by integrating:

\(\int_{1}^{5} (6x - x^2) \, dx = \left[ 3x^2 - \frac{1}{3}x^3 \right]_{1}^{5}\)

Calculating the definite integral:

\(\left( 3(5)^2 - \frac{1}{3}(5)^3 \right) - \left( 3(1)^2 - \frac{1}{3}(1)^3 \right)\)

\(= (75 - \frac{125}{3}) - (3 - \frac{1}{3})\)

\(= \frac{225}{3} - \frac{125}{3} - \left( \frac{9}{3} - \frac{1}{3} \right)\)

\(= \frac{100}{3} - \frac{8}{3}\)

\(= \frac{92}{3}\)

The area of the rectangle formed by the line \(y = 5\) from \(x = 1\) to \(x = 5\) is:

\(5 \times (5 - 1) = 20\)

The shaded area is the area under the curve minus the area of the rectangle:

\(\frac{92}{3} - 20 = \frac{92}{3} - \frac{60}{3} = \frac{32}{3}\)

Thus, the shaded area is \(\frac{10}{3}\).

To find the area of the shaded region, we need to integrate the function \(y = 10x^{\frac{1}{2}} - \frac{5}{2}x^{\frac{3}{2}}\) from \(x = 0\) to \(x = 4\).

The integral is:

\(\int_{0}^{4} \left( 10x^{\frac{1}{2}} - \frac{5}{2}x^{\frac{3}{2}} \right) \, dx\)

Integrating term by term, we have:

\(\int 10x^{\frac{1}{2}} \, dx = \frac{10}{\frac{3}{2}} x^{\frac{3}{2}} = \frac{20}{3} x^{\frac{3}{2}}\)

\(\int -\frac{5}{2}x^{\frac{3}{2}} \, dx = -\frac{5}{2} \cdot \frac{2}{5} x^{\frac{5}{2}} = -x^{\frac{5}{2}}\)

Thus, the integral becomes:

\(\left[ \frac{20}{3} x^{\frac{3}{2}} - x^{\frac{5}{2}} \right]_{0}^{4}\)

Evaluating at the bounds:

\(\left( \frac{20}{3} \times 8 - 32 \right) - \left( 0 \right) = \frac{160}{3} - 32\)

\(= \frac{160}{3} - \frac{96}{3} = \frac{64}{3}\)

Therefore, the area of the shaded region is \(\frac{64}{3}\) or approximately \(21.3\).