First, find the points of intersection by setting the equations equal: \(2x + 1 = -x^2 + 12x - 20\).

Rearrange to get: \(x^2 - 10x + 21 = 0\).

Factorize to find \((x - 3)(x - 7) = 0\), so \(x = 3\) and \(x = 7\).

Calculate the area under the line \(y = 2x + 1\) from \(x = 3\) to \(x = 7\):

Area of trapezium = \(\frac{1}{2} \times 4 \times (7 + 15) = 44\).

Calculate the area under the curve \(y = -x^2 + 12x - 20\) from \(x = 3\) to \(x = 7\):

Integrate to find \(\int_{3}^{7} (-x^2 + 12x - 20) \, dx = \left[ -\frac{x^3}{3} + 6x^2 - 20x \right]_{3}^{7}\).

Evaluate the integral: \(\left[ -\frac{343}{3} + 294 - 140 \right] - \left[ -\frac{27}{3} + 54 - 60 \right] = 54\frac{2}{3}\).

The shaded area is the difference: \(44 - 54\frac{2}{3} = 10\frac{2}{3}\).

(i) To find \(\frac{dy}{dx}\), differentiate \(y = 8 - \sqrt{4-x}\). Using the chain rule, \(\frac{dy}{dx} = -\frac{1}{2}(4-x)^{-\frac{1}{2}} \times (-1)\).

For \(\int y \, dx\), integrate \(y = 8 - \sqrt{4-x}\). The integral is \(8x - \frac{(4-x)^{\frac{3}{2}}}{\frac{3}{2}} \times (-1)\).

(ii) The equation of the tangent at \(P(3, 7)\) is found using the point-slope form. The slope \(m = \frac{1}{2}\), so the equation is \(y - 7 = \frac{1}{2}(x - 3)\), which simplifies to \(y = \frac{1}{2}x + \frac{5}{2}\).

(iii) The area under the curve from 0 to 3 is \(\int_{0}^{3} (8 - \sqrt{4-x}) \, dx = \frac{58}{3}\). The area under the line is \(\frac{1}{2}(\frac{5}{2} + 7) \times 3 = \frac{75}{4}\). The area of the shaded region is \(\frac{58}{3} - \frac{75}{4} = \frac{7}{12}\).

To find the points of intersection, set the equations equal: \(2x + 11 = 8 - 2x - x^2\).

Rearrange to form a quadratic equation: \(x^2 + 4x + 3 = 0\).

Factorize: \((x + 1)(x + 3) = 0\).

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

To find the area between the curves, integrate the difference between the curve and the line from \(x = -3\) to \(x = -1\):

\(\int_{-3}^{-1} (8 - 2x - x^2) \, dx - \int_{-3}^{-1} (2x + 11) \, dx\).

Integrate each part:

\(\int (8 - 2x - x^2) \, dx = 8x - x^2 - \frac{x^3}{3}\).

\(\int (2x + 11) \, dx = x^2 + 11x\).

Evaluate from \(x = -3\) to \(x = -1\):

\([8(-1) - (-1)^2 - \frac{(-1)^3}{3}] - [8(-3) - (-3)^2 - \frac{(-3)^3}{3}]\).

\([8(-1) - 1 + \frac{1}{3}] - [8(-3) - 9 + \frac{27}{3}]\).

Calculate the definite integrals and subtract: \(\frac{4}{3}\).

(a) To find the maximum point, differentiate the curve equation: \(\frac{dy}{dx} = 9 - \frac{3}{2}(2x+1)^{\frac{1}{2}} \times 2\).

Set \(\frac{dy}{dx} = 0\):

\(9 - 3(2x+1)^{\frac{1}{2}} = 0\)

\(2x+1 = 9 \Rightarrow x = 4\)

Substitute \(x = 4\) into the original equation to find \(y\):

\(y = 9(4) - (2(4) + 1)^{\frac{3}{2}} = 36 - 9 = 27\)

Thus, the maximum point is \((4, 9)\).

(b) The gradient of the curve at \(A\) is found by substituting \(x = 1\frac{1}{2}\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = 3\)

The gradient of \(AB\) is:

\(\frac{5\frac{1}{2} - 3\frac{1}{2}}{1\frac{1}{2} - 7\frac{1}{2}} = \frac{2}{-6} = -\frac{1}{3}\)

Since the product of the gradients is \(-1\), \(AB\) is the normal to the curve at \(A\).

(c) To find the area of the shaded region, integrate the curve equation from \(x = 1\frac{1}{2}\) to \(x = 7\frac{1}{2}\):

\(\int_{1\frac{1}{2}}^{7\frac{1}{2}} \left(9x - (2x+1)^{\frac{3}{2}}\right) \, dx\)

Calculate the definite integral and subtract the area of the trapezium formed by \(AB\):

\(Shaded area = 44.6 - 27 = 17.6\)

(i) To find the equation of the tangent, we first need the derivative of the curve \(y = (3 - 2x)^3\).

Using the chain rule, \(\frac{dy}{dx} = 3(3 - 2x)^2 \times (-2) = -6(3 - 2x)^2\).

At \(x = \frac{1}{2}\), \(\frac{dy}{dx} = -6(3 - 2 \times \frac{1}{2})^2 = -6(2)^2 = -24\).

The slope of the tangent is \(-24\). Using the point-slope form, \(y - 8 = -24(x - \frac{1}{2})\).

Simplifying gives \(y = -24x + 20\).

(ii) To find the area of the shaded region, calculate the area under the curve and subtract the area under the tangent.

Area under the curve from \(x = 0\) to \(x = \frac{1}{2}\):

\(\int_0^{1/2} (3 - 2x)^3 \, dx = \left[ \frac{(3 - 2x)^4}{-8} \right]_0^{1/2} = \left[ \frac{(2)^4}{-8} - \frac{(3)^4}{-8} \right] = \frac{81}{8}\).

Area under the tangent from \(x = 0\) to \(x = \frac{1}{2}\):

\(\int_0^{1/2} (-24x + 20) \, dx = \left[ -12x^2 + 20x \right]_0^{1/2} = -3 + 10 = 7\).

Shaded area = \(\frac{81}{8} - 7 = \frac{9}{8}\) or 1.125.