(i) Substitute \(x = -2\) and \(x = -3\) into the function:

\(2a + 4b = 8\)

\(2a^2 + 3a + 4b = 14\)

Substitute \(4b = 8 - 2a\) into the second equation:

\(2a^2 + 3a + (8 - 2a) = 14\)

\(2a^2 + a - 6 = 0\)

\((a + 2)(2a - 3) = 0\)

Thus, \(a = -2\) or \(a = \frac{3}{2}\).

For \(a = -2\), \(2(-2) + 4b = 8\) gives \(b = 3\).

For \(a = \frac{3}{2}\), \(2(\frac{3}{2}) + 4b = 8\) gives \(b = \frac{5}{4}\).

(ii) For \(a = 1\) and \(b = -1\), the function is \(f(x) = x^2 - x - 3\).

Complete the square: \(y = \left(x - \frac{1}{2}\right)^2 - \frac{13}{4}\).

Rearrange to find \(x\):

\(x - \frac{1}{2} = \pm \sqrt{y + \frac{13}{4}}\)

\(x = \frac{1}{2} \pm \sqrt{y + \frac{13}{4}}\)

Choose the positive root for the inverse: \(f^{-1}(x) = \frac{1}{2} \sqrt{x + \frac{13}{4}}\).

The domain of \(f^{-1}\) is \(x \geq -\frac{13}{4}\).