(i) To find the inverse function \(f^{-1}(x)\), start with \(y = 4 - 5x\).

Solve for \(x\):

\(y = 4 - 5x\)

\(5x = 4 - y\)

\(x = \frac{4-y}{5}\)

Thus, \(f^{-1}(x) = \frac{4-x}{5}\).

To find the intersection, set \(f(x) = f^{-1}(x)\):

\(4 - 5x = \frac{4-x}{5}\)

Multiply through by 5:

\(20 - 25x = 4 - x\)

\(20 - 4 = 25x - x\)

\(16 = 24x\)

\(x = \frac{2}{3}\)

Substitute back to find \(y\):

\(y = 4 - 5 \left( \frac{2}{3} \right) = \frac{2}{3}\)

So, the point of intersection is \(\left( \frac{2}{3}, \frac{2}{3} \right)\).

(ii) The graph of \(y = f(x) = 4 - 5x\) is a straight line with a negative slope, intersecting the y-axis at 4.

The graph of \(y = f^{-1}(x) = \frac{4-x}{5}\) is also a straight line with a negative slope, intersecting the y-axis at \(\frac{4}{5}\).

The line \(y = x\) is the line of symmetry between the function and its inverse.

The graph of the inverse function \(y = f^{-1}(x)\) is obtained by reflecting the graph of \(y = f(x)\) across the line \(y = x\). This means that for every point \((a, b)\) on the graph of \(y = f(x)\), there is a corresponding point \((b, a)\) on the graph of \(y = f^{-1}(x)\).

In the diagram, the curve \(y = f(x)\) is reflected over the line \(y = x\) to produce the curve \(y = f^{-1}(x)\).

(ii) For \(g(x) = 8 - (x - 2)^2\) to have an inverse, it must be one-to-one. The function is a downward-opening parabola with vertex at \(x = 2\). To ensure it is one-to-one, we restrict the domain to \(x \geq 2\). Therefore, the smallest value of \(k\) is 2.

(iii) To find \(g^{-1}(x)\), start with \(y = 8 - (x - 2)^2\).

Rearrange to make \(x\) the subject:

\((x - 2)^2 = 8 - y\)

\(x - 2 = \pm \sqrt{8 - y}\)

Since \(x \geq 2\), we take the positive root:

\(x = 2 + \sqrt{8 - y}\)

Thus, \(g^{-1}(x) = 2 + \sqrt{8 - x}\).

(iv) The sketch should show the graph of \(y = g(x)\) as a downward-opening parabola starting at \(x = 2\) and ending at \(x = 4\). The graph of \(y = g^{-1}(x)\) should be the reflection of \(y = g(x)\) across the line \(y = x\), starting at \(y = 2\) and ending at \(y = 4\).

(i) To express \(f(x) = 2x^2 - 8x + 10\) in the form \(a(x+b)^2 + c\), complete the square:

\(f(x) = 2(x^2 - 4x) + 10\)

\(= 2((x-2)^2 - 4) + 10\)

\(= 2(x-2)^2 - 8 + 10\)

\(= 2(x-2)^2 + 2\)

(ii) The vertex form \(2(x-2)^2 + 2\) shows the minimum value is 2 when \(x = 2\), and the maximum value is 10 when \(x = 0\). Thus, the range is \(2 \leq f(x) \leq 10\).

(iii) The range of \(f\) is \(2 \leq f(x) \leq 10\), so the domain of \(f^{-1}\) is \(2 \leq x \leq 10\).

(iv) The graph of \(f(x)\) is a half parabola from (0,10) to (2,2). The graph of \(g(x)\) is a line through the origin at 45°. The graph of \(f^{-1}(x)\) is the reflection of \(f(x)\) in \(g(x)\).

(v) To find \(f^{-1}(x)\), start from \(y = 2(x-2)^2 + 2\):

\((x-2)^2 = \frac{1}{2}(y-2)\)

\(x = 2 \pm \sqrt{\frac{1}{2}(y-2)}\)

Since \(f(x)\) is decreasing on \(0 \leq x \leq 2\), choose the negative root:

\(f^{-1}(x) = 2 - \sqrt{\frac{1}{2}(x-2)}\)

(ii) Start with \(y = \frac{6}{2x+3}\). To find \(f^{-1}(x)\), interchange \(x\) and \(y\):

\(x = \frac{6}{2y+3}\)

Solve for \(y\):

\(x(2y+3) = 6\)

\(2xy + 3x = 6\)

\(2xy = 6 - 3x\)

\(y = \frac{6 - 3x}{2x}\)

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

The domain of \(f^{-1}\) is \(0 < x \leq 2\) because \(f(x)\) is defined for \(x \geq 0\) and \(f(x) \to 0\) as \(x \to \infty\).

(iii) The graph of \(y = f^{-1}(x)\) is a reflection of \(y = f(x)\) across the line \(y = x\).

(iv) Given \(fg(x) = \frac{3}{2}\), substitute \(g(x) = \frac{1}{2}x\):

\(f\left(\frac{1}{2}x\right) = \frac{3}{2}\)

\(\frac{6}{2\left(\frac{1}{2}x\right) + 3} = \frac{3}{2}\)

\(\frac{6}{x + 3} = \frac{3}{2}\)

Cross-multiply:

\(12 = 3(x + 3)\)

\(12 = 3x + 9\)

\(3 = 3x\)

\(x = 1\)