(i) The curve is given by \(y = x(x-1)(x-2) = x^3 - 3x^2 + 2x\).

Differentiate to find the gradient: \(\frac{dy}{dx} = 3x^2 - 6x + 2\).

At \(A(1, 0)\), the gradient \(m = -1\), so the tangent is \(y = -1(x-1)\).

At \(B(2, 0)\), the gradient \(m = 2\), so the tangent is \(y = 2(x-2)\).

Solving the equations \(y = -x + 1\) and \(y = 2x - 4\) gives \(x = \frac{5}{3}\).

(ii) The area \(R_1\) is given by \(\int_0^1 (x^3 - 3x^2 + 2x) \, dx\).

Integrate: \(\left[ \frac{x^4}{4} - \frac{3x^3}{3} + \frac{2x^2}{2} \right]_0^1 = \frac{1}{4}\).

The area \(R_2\) is given by \(\int_1^2 (x^3 - 3x^2 + 2x) \, dx\).

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

Thus, \(R_1 = R_2 = \frac{1}{4}\).

(i) Differentiate the curve: \(\frac{dy}{dx} = 3x^2 - 6x - 9\).

Set \(\frac{dy}{dx} = 0\) to find critical points: \(3x^2 - 6x - 9 = 0\).

Solving gives \(x = 3\) or \(x = -1\).

Since the minimum point is on the \(x\)-axis, \(y = 0\) at \(x = 3\).

Substitute \(x = 3\) into the curve equation: \(0 = 3^3 - 3(3)^2 - 9(3) + k\).

Simplify: \(0 = 27 - 27 - 27 + k\).

Thus, \(k = 27\).

(iv) Integrate \(y\) to find the area: \(\int (x^3 - 3x^2 - 9x + 27) \, dx\).

The integral is \(\frac{x^4}{4} - x^3 - \frac{9x^2}{2} + 27x\).

Evaluate from \(x = 0\) to \(x = 3\):

\(\left[ \frac{3^4}{4} - 3^3 - \frac{9(3)^2}{2} + 27(3) \right] - \left[ \frac{0^4}{4} - 0^3 - \frac{9(0)^2}{2} + 27(0) \right]\).

Calculate: \(\frac{81}{4} - 27 - 40.5 + 81 = 33.75\).

Thus, the area of the shaded region is 33.75.

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = \frac{16}{x^3}\):

\(y = \int \frac{16}{x^3} \, dx = \int 16x^{-3} \, dx = \frac{16x^{-2}}{-2} + c = \frac{-8}{x^2} + c\).

Using the point \((1, 4)\), substitute to find \(c\):

\(4 = \frac{-8}{1^2} + c \Rightarrow c = 12\).

Thus, the equation of the curve is \(y = \frac{-8}{x^2} + 12\).

(ii) The normal has gradient \(-\frac{1}{2}\), so the perpendicular gradient is 2. Set \(\frac{16}{x^3} = 2\) to find \(x\):

\(16 = 2x^3 \Rightarrow x = 2\).

Substitute \(x = 2\) into the curve equation to find \(y\):

\(y = \frac{-8}{2^2} + 12 = 10\).

The equation of the normal is \(y - 10 = -\frac{1}{2}(x - 2)\).

Simplify to \(2y + x = 22\).

(iii) Find the area under the curve from \(x = 1\) to \(x = 2\):

\(A = \int_{1}^{2} \left( \frac{-8}{x^2} + 12 \right) \, dx\).

\(= \left[ \frac{8}{x} + 12x \right]_{1}^{2}\).

\(= \left( \frac{8}{2} + 12 \times 2 \right) - \left( \frac{8}{1} + 12 \times 1 \right)\).

\(= (4 + 24) - (8 + 12) = 28 - 20 = 8\).

To find the area under the curve \(y = \frac{4}{\sqrt{x}}\) from \(x = 1\) to \(x = 4\), we need to evaluate the definite integral:

\(\int_{1}^{4} 4x^{-0.5} \, dx\)

First, find the antiderivative of \(4x^{-0.5}\):

\(\int 4x^{-0.5} \, dx = \frac{4x^{0.5}}{0.5} = 8x^{0.5}\)

Now, evaluate this antiderivative from 1 to 4:

\(\left[ 8x^{0.5} \right]_{1}^{4} = 8 \sqrt{4} - 8 \sqrt{1}\)

\(= 8 \times 2 - 8 \times 1\)

\(= 16 - 8 = 8\)

Thus, the area of the region is 8.

To find the area enclosed by the curve \(y = \sqrt{5x + 4}\), the \(x\)-axis, the \(y\)-axis, and the line \(x = 1\), we need to integrate the function from \(x = 0\) to \(x = 1\).

The integral to evaluate is:

\(\int_0^1 \sqrt{5x + 4} \, dx\)

First, rewrite \(\sqrt{5x + 4}\) as \((5x + 4)^{1/2}\).

Using the power rule for integration, the integral becomes:

\(\int (5x + 4)^{1/2} \, dx = \frac{(5x + 4)^{3/2}}{\frac{3}{2} \cdot 5} = \frac{2}{15} (5x + 4)^{3/2}\)

Evaluate this from \(x = 0\) to \(x = 1\):

\(\left[ \frac{2}{15} (5x + 4)^{3/2} \right]_0^1 = \frac{2}{15} ((5 \cdot 1 + 4)^{3/2} - (5 \cdot 0 + 4)^{3/2})\)

\(= \frac{2}{15} (9^{3/2} - 4^{3/2})\)

\(9^{3/2} = 27\) and \(4^{3/2} = 8\), so:

\(\frac{2}{15} (27 - 8) = \frac{2}{15} \cdot 19 = \frac{38}{15} \approx 2.53\)