(i) To find the stationary point, we first find the derivative of the curve: \(\frac{dy}{dx} = 4x^3 + 4\). Setting this to zero gives \(4x^3 + 4 = 0\), which simplifies to \(x^3 = -1\), so \(x = -1\). Substituting \(x = -1\) back into the original equation gives \(y = (-1)^4 + 4(-1) + 9 = 1 - 4 + 9 = 6\). Thus, the stationary point is \((-1, 6)\).

To determine the nature, we find the second derivative: \(\frac{d^2y}{dx^2} = 12x^2\). Substituting \(x = -1\) gives \(12(-1)^2 = 12\), which is positive, indicating a minimum point.

(ii) To find the area, we integrate the curve from \(x = 0\) to \(x = 1\). The integral of \(y = x^4 + 4x + 9\) is \(\int (x^4 + 4x + 9) \, dx = \frac{x^5}{5} + 2x^2 + 9x\). Evaluating from 0 to 1 gives:

\(\left[ \frac{x^5}{5} + 2x^2 + 9x \right]_0^1 = \left( \frac{1^5}{5} + 2(1)^2 + 9(1) \right) - \left( \frac{0^5}{5} + 2(0)^2 + 9(0) \right) = \frac{1}{5} + 2 + 9 = 11.2\).

(i) To find the coordinates of \(A\) and \(B\), we first find the derivative \(\frac{dy}{dx} = 3x^2 - 12x + 9\). Setting \(\frac{dy}{dx} = 0\) gives the critical points. Solving \(3x^2 - 12x + 9 = 0\) gives \(x = 1\) and \(x = 3\). Substituting back into the original equation, \(y = x^3 - 6x^2 + 9x\), we find \(A(1, 4)\) and \(B(3, 0)\).

(ii) At \(C(2, 2)\), the slope of the tangent is \(m = -3\). The slope of the normal is \(\frac{1}{3}\). The equation of the normal is \(y - 2 = \frac{1}{3}(x - 2)\), which simplifies to \(3y = x + 4\).

(iii) To find the area of the shaded region, integrate \(y = x^3 - 6x^2 + 9x\) from \(x = 0\) to \(x = 3\). The integral is \(\frac{1}{4}x^4 - 2x^3 + \frac{9}{2}x^2\). Evaluating from 0 to 3 gives \(\frac{3}{4}\). The area of the trapezium is \(\frac{1}{2} \times 1 \times (2 + \frac{2}{3}) = \frac{13}{6}\). Subtracting gives the shaded area \(\frac{17}{12}\).

(i) Integrate \(\frac{dy}{dx} = -\frac{k}{x^3}\) to get \(y = -k \frac{x^{-2}}{-2} + c = \frac{k}{2x^2} + c\).

Substitute \((1, 18)\):

\(18 = \frac{k}{2} + c\)

Substitute \((4, 3)\):

\(3 = \frac{k}{32} + c\)

Solving these equations gives \(k = 32\) and \(c = 2\).

Thus, \(y = \frac{16}{x^2} + 2\).

(ii) The area of the shaded region is given by:

\(\int_{1}^{1.6} \left( \frac{16}{x^2} + 2 \right) \, dx\)

Calculate:

\(\left[ -\frac{16}{x} + 2x \right]_{1}^{1.6} = \left[ -10 + 3.2 \right] - \left[ -16 + 2 \right] = 7.2\).

The area under the curve \(y = 2\sqrt{x}\) from \(x = 1\) to \(x = 4\) is given by the definite integral:

\(\int_{1}^{4} 2\sqrt{x} \, dx\)

First, rewrite \(2\sqrt{x}\) as \(2x^{0.5}\).

The integral of \(2x^{0.5}\) is:

\(\frac{2}{1.5} x^{1.5} = \frac{4}{3} x^{1.5}\)

Evaluate this from \(x = 1\) to \(x = 4\):

\(\left[ \frac{4}{3} x^{1.5} \right]_{1}^{4} = \frac{4}{3} (4^{1.5}) - \frac{4}{3} (1^{1.5})\)

Calculate \(4^{1.5} = 8\) and \(1^{1.5} = 1\):

\(\frac{4}{3} (8) - \frac{4}{3} (1) = \frac{32}{3} - \frac{4}{3} = \frac{28}{3}\)

The area under the curve from \(x = 1\) to \(x = 2\) is given by the integral:

\(\int_{1}^{2} \left( 2x + \frac{8}{x^2} \right) \, dx\)

We can split this into two separate integrals:

\(\int_{1}^{2} 2x \, dx + \int_{1}^{2} \frac{8}{x^2} \, dx\)

The first integral is:

\(\int 2x \, dx = x^2\)

Evaluated from 1 to 2:

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

The second integral is:

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

Evaluated from 1 to 2:

\(\left[-\frac{8}{x}\right]_{1}^{2} = -\frac{8}{2} + \frac{8}{1} = -4 + 8 = 4\)

Adding both results gives the total area:

\(3 + 4 = 7\)