(i) At \(x = 2\), the gradient \(m = k - 4\). At \(x = 3\), the gradient \(m = k - 6\). Since the tangents are perpendicular, \((k - 4)(k - 6) = -1\).

Solving \((k - 4)(k - 6) = -1\):

\(k^2 - 10k + 24 = -1\)

\(k^2 - 10k + 25 = 0\)

\((k - 5)^2 = 0\)

\(k = 5\)

(ii) Integrate \(\frac{dy}{dx} = 5 - 2x\) to find \(y\):

\(y = 5x - x^2 + c\)

Substitute the point (4, 9):

\(9 = 5(4) - 4^2 + c\)

\(9 = 20 - 16 + c\)

\(c = 5\)

Thus, the equation of the curve is \(y = 5x - x^2\).

(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = 4 - x\):

\(y = \int (4 - x) \, dx = 4x - \frac{1}{2}x^2 + c\).

Using the point \(P(2, 9)\), substitute to find \(c\):

\(9 = 4(2) - \frac{1}{2}(2)^2 + c\)

\(9 = 8 - 2 + c\)

\(c = 3\).

Thus, the equation of the curve is \(y = 4x - \frac{1}{2}x^2 + 3\).

(ii) The gradient of the tangent at \(P\) is \(\frac{dy}{dx} = 4 - 2 = 2\).

The gradient of the normal is \(-\frac{1}{2}\) (since \(m_1 m_2 = -1\)).

The equation of the normal is \(y - 9 = -\frac{1}{2}(x - 2)\).

(iii) Substitute \(y = 4x - \frac{1}{2}x^2 + 3\) into the normal equation:

\(4x - \frac{1}{2}x^2 + 3 - 9 = -\frac{1}{2}(x - 2)\)

\(4x - \frac{1}{2}x^2 - 6 = -\frac{1}{2}x + 1\)

\(x^2 - 9x + 14 = 0\)

Solving the quadratic equation gives \(x = 7\).

Substitute \(x = 7\) back into the curve equation:

\(y = 4(7) - \frac{1}{2}(7)^2 + 3\)

\(y = 28 - 24.5 + 3\)

\(y = 6.5\).

Thus, the coordinates of \(Q\) are \((7, 6.5)\).

(i) At \(x = 1\), the gradient \(m\) of the tangent is \(2\). The gradient of the normal is \(-\frac{1}{2}\) since \(m \cdot m_\text{normal} = -1\).

The equation of the normal is \(y - 8 = -\frac{1}{2}(x - 1)\), which simplifies to \(2y + x = 17\).

The normal meets the x-axis when \(y = 0\), giving \(x = 17\), so \(R(17, 0)\).

The normal meets the y-axis when \(x = 0\), giving \(y = 8.5\), so \(Q(0, 8.5)\).

The mid-point of \(QR\) is \(\left( \frac{0 + 17}{2}, \frac{8.5 + 0}{2} \right) = \left( \frac{17}{2}, \frac{9}{2} \right)\).

(ii) Integrate \(\frac{dy}{dx} = \frac{4}{\sqrt{6 - 2x}}\) to find \(y\).

\(y = \int \frac{4}{\sqrt{6 - 2x}} \, dx = 4(6 - 2x)^{\frac{1}{2}} \cdot \left(-\frac{1}{2}\right) + c\).

\(y = \frac{1}{2}x - 2 + c\).

Substitute \(P(1, 8)\) into the equation: \(8 = \frac{1}{2}(1) - 2 + c\).

Solving gives \(c = 16\).

Thus, the equation of the curve is \(y = \frac{1}{2}x - 2 + 16\).

(i) To find where \(f\) is decreasing, we need \(f'(x) < 0\).

\(3(2x-1)^{\frac{1}{2}} - 6 < 0\)

\(3(2x-1)^{\frac{1}{2}} < 6\)

\((2x-1)^{\frac{1}{2}} < 2\)

\(2x-1 < 4\)

\(2x < 5\)

\(x < \frac{5}{2}\)

Since \(x > \frac{1}{2}\) is given, the set of values is \(\frac{1}{2} < x < \frac{5}{2}\).

(ii) We integrate \(f'(x) = 3(2x-1)^{\frac{1}{2}} - 6\) to find \(f(x)\).

\(f(x) = \int (3(2x-1)^{\frac{1}{2}} - 6) \, dx\)

\(f(x) = \left[ \frac{3}{2} (2x-1)^{\frac{3}{2}} \cdot \frac{1}{2} \right] - 6x + c\)

\(f(x) = (2x-1)^{\frac{3}{2}} - 6x + c\)

Given \(f(1) = -3\), substitute \(x = 1\):

\(-3 = (2 \cdot 1 - 1)^{\frac{3}{2}} - 6 \cdot 1 + c\)

\(-3 = 1 - 6 + c\)

\(c = 2\)

Thus, \(f(x) = (2x-1)^{\frac{3}{2}} - 6x + 2\).

(i) To find the equation of the normal, first find the gradient of the tangent at \(P(3, 3)\). The gradient \(m\) is given by \(\frac{dy}{dx} = \frac{6}{\sqrt{4x - 3}}\). At \(x = 3\), \(\frac{dy}{dx} = \frac{6}{\sqrt{4(3) - 3}} = 2\). The gradient of the normal is the negative reciprocal, \(m = -\frac{1}{2}\). The equation of the normal is \(y - 3 = -\frac{1}{2}(x - 3)\), which simplifies to \(x + 2y = 9\).

(ii) To find the equation of the curve, integrate \(\frac{dy}{dx} = \frac{6}{\sqrt{4x - 3}}\). Let \(u = 4x - 3\), then \(du = 4dx\), so \(dx = \frac{1}{4}du\). The integral becomes \(\int \frac{6}{\sqrt{u}} \cdot \frac{1}{4} \, du = \frac{3}{2} \int u^{-1/2} \, du\). This integrates to \(3u^{1/2} + c\). Substituting back \(u = 4x - 3\), we get \(y = 3(4x - 3)^{1/2} + c\). Using the point \((3, 3)\), substitute to find \(c\): \(3 = 3(4(3) - 3)^{1/2} + c\), which gives \(c = -6\). Therefore, the equation of the curve is \(y = 3(4x - 3)^{1/2} - 6\).