(ii) To find the distance \(AB\), use the distance formula:

\(AB^2 = \left(3/2\right)^2 + \left(3/4\right)^2\)

\(AB = \sqrt{\frac{45}{4}} \approx 1.68\)

(iii) To find the area of the shaded region, calculate the area under the curve \(y = 2x + (x+1)^{-2}\) from \(x = -\frac{1}{2}\) to \(x = 1\).

The area under the curve is:

\(\int_{-1/2}^{1} \left( x^2 - (x+1)^{-1} \right) \, dx = \left[ \frac{1}{4}x^2 + \frac{11}{4}x \right]_{-1/2}^{1} - \left[ x^2 - (x+1)^{-1} \right]_{-1/2}^{1}\)

Apply limits \(-\frac{1}{2} \to 1\) to both integrals:

\(= \frac{27}{16} \approx 1.69\)

(i) To find the equation of the tangent, we first find the derivative of the curve \(y = 9 + 6x - 3x^2\), which is \(\frac{dy}{dx} = 6 - 6x\). At \(x = 2\), the gradient is \(6 - 6(2) = -6\). The equation of the tangent at A (2, 9) is \(y - 9 = -6(x - 2)\), which simplifies to \(y = -6x + 21\). To find the x-coordinate of C, set \(y = 0\) in the tangent equation: \(0 = -6x + 21\), giving \(x = \frac{21}{6} = \frac{7}{2}\).

(ii) The area under the curve from \(x = 2\) to \(x = 3\) is calculated by integrating \(9 + 6x - 3x^2\), which gives \(9x + 3x^2 - x^3\). Evaluating from 2 to 3, we get \((27 + 27 - 27) - (18 + 12 - 8) = 5\). The area under the tangent from \(x = 2\) to \(x = \frac{7}{2}\) is \(\frac{1}{2} \times \frac{3}{2} \times 9 = \frac{27}{4}\). The area of the shaded region ABC is \(\frac{27}{4} - 5 = \frac{7}{4}\).

(i) The equation of the curve is \(y = \frac{8}{\sqrt{3x+4}}\). The derivative is \(\frac{dy}{dx} = \frac{-4}{(3x+4)^{\frac{3}{2}}} \times 3\). At \(x = 0\), \(\frac{dy}{dx} = -\frac{3}{2}\). The slope of the normal is the negative reciprocal, \(\frac{2}{3}\).

The equation of the normal at \(A\) is \(y - 4 = \frac{2}{3}(x - 0)\). Solving for \(x = 4\), we find \(B \left( 4, \frac{20}{3} \right)\).

(ii) The area \(P\) is given by \(\int_{0}^{4} \frac{8}{\sqrt{3x+4}} \, dx = 8 \int_{0}^{4} (3x+4)^{-\frac{1}{2}} \, dx\). This evaluates to \(\frac{32}{3}\).

The area \(Q\) is the area of the trapezium minus \(P\). The area of the trapezium is \(\frac{1}{2} \times (4 + \frac{20}{3}) \times 4 = \frac{64}{3}\). Thus, \(Q = \frac{64}{3} - \frac{32}{3} = \frac{32}{3}\).

Therefore, the areas of \(P\) and \(Q\) are both \(\frac{32}{3}\).

(i) Set the equations equal: \(x + 2 = 3\sqrt{x}\).

Rearrange: \(x - 3\sqrt{x} + 2 = 0\).

Let \(k = \sqrt{x}\), then \(k^2 - 3k + 2 = 0\).

Factorize: \((k-1)(k-2) = 0\).

So, \(k = 1\) or \(k = 2\).

Thus, \(\sqrt{x} = 1\) or \(\sqrt{x} = 2\).

Therefore, \(x = 1\) or \(x = 4\).

For \(x = 1\), \(y = 3\).

For \(x = 4\), \(y = 6\).

Coordinates of \(A\) are \((1, 3)\) and \(B\) are \((4, 6)\).

(ii) Find the area between the curves from \(x = 1\) to \(x = 4\).

Area = \(\int_1^4 (3\sqrt{x} - (x+2)) \, dx\).

Integrate: \(\int_1^4 (3x^{1/2} - x - 2) \, dx\).

\(= \left[ 2x^{3/2} - \frac{1}{2}x^2 - 2x \right]_1^4\).

Calculate: \(\left[ 16 - 8 - 8 \right] - \left[ 2 - \frac{1}{2} - 2 \right]\).

\(= 0 - (-1) = 1\).

Area = \(\frac{1}{2}\).

(i) For \(y = (4x + 1)^{\frac{1}{2}}\), the derivative is \(\frac{dy}{dx} = \frac{1}{2}(4x + 1)^{-\frac{1}{2}} \times 4 = \frac{2}{\sqrt{4x + 1}}\).

When \(x = 2\), \(\frac{dy}{dx} = \frac{2}{3}\).

For \(y = \frac{1}{2}x^2 + 1\), the derivative is \(\frac{dy}{dx} = x\).

When \(x = 2\), \(\frac{dy}{dx} = 2\).

The angle \(\alpha\) is given by \(\alpha = \arctan(m_2) - \arctan(m_1)\), where \(m_1 = \frac{2}{3}\) and \(m_2 = 2\).

\(\alpha = \arctan(2) - \arctan\left(\frac{2}{3}\right) \approx 63.43 - 33.69 = 29.7\) degrees.

(ii) The area of the shaded region is found by integrating the difference between the two curves from \(x = 0\) to \(x = 2\).

\(\int (4x + 1)^{\frac{1}{2}} \, dx = \left[ \frac{(4x+1)^{\frac{3}{2}}}{\frac{3}{2} \times 4} \right] = \left[ \frac{(4x+1)^{\frac{3}{2}}}{6} \right]\).

\(\int \left(\frac{1}{2}x^2 + 1\right) \, dx = \frac{1}{6}x^3 + x\).

Evaluate from 0 to 2:

\(\left[ \frac{(4 \times 2 + 1)^{\frac{3}{2}}}{6} \right] - \left[ \frac{(4 \times 0 + 1)^{\frac{3}{2}}}{6} \right] = \frac{27}{6} - \frac{1}{6} = \frac{26}{6} = \frac{13}{3}\).

\(\left[ \frac{1}{6}(2)^3 + 2 \right] - \left[ \frac{1}{6}(0)^3 + 0 \right] = \frac{8}{6} + 2 = \frac{10}{3}\).

The area is \(\frac{13}{3} - \frac{10}{3} = 1\).