(i) Given \(fg(x) = f(x) \cdot g(x) = 6x^2 - 21\), substitute \(f(x) = 2x + 3\) and \(g(x) = ax^2 + b\):

\(2(ax^2 + b) + 3 = 6x^2 - 21\)

\(2ax^2 + 2b + 3 = 6x^2 - 21\)

Equating coefficients, \(2a = 6\) gives \(a = 3\), and \(2b + 3 = -21\) gives \(b = -12\).

(ii) For \(fg(x) = 6x^2 - 21\) to be defined, \(x \leq q\). Solve \(6x^2 - 21 \geq 3\):

\(6x^2 \geq 24\)

\(x^2 \geq 4\)

\(x \leq -2\) or \(x \geq 2\). Since \(x \leq q\), the greatest possible \(q = -2\).

(iii) Given \(q = -3\), find the range of \(fg\):

\(y = 6(-3)^2 - 21 = 33\)

Thus, the range is \(y \geq 33\).

(iv) Solve \(y = 6x^2 - 21\) for \(x\):

\(y + 21 = 6x^2\)

\(x^2 = \frac{y + 21}{6}\)

\(x = \pm \sqrt{\frac{y + 21}{6}}\)

Since \(x \leq q\), choose the negative root: \(x = -\sqrt{\frac{y + 21}{6}}\)

Thus, \((fg)^{-1}(x) = -\sqrt{\frac{x + 21}{6}}\).

The domain of \((fg)^{-1}\) is \(x \geq 33\).