(a)(i) The function \(f(x) = (x - 3)^2 - 1\) is defined for \(x < a\). The greatest possible value of \(a\) is 3, as the function is one-one up to this point.

(a)(ii) For \(a = 3\), the range of \(f\) is \(y > -1\). To find \(f^{-1}(x)\), set \(y = (x - 3)^2 - 1\), so \(y + 1 = (x - 3)^2\). Solving for \(x\), we get \(x = 3 \pm \sqrt{1 + y}\). Since \(x < 3\), we take \(x = 3 - \sqrt{1 + y}\), hence \(f^{-1}(x) = 3 - \sqrt{1 + x}\).

(b)(i) \(gg(2x) = [(2x - 3)^2 - 3]^2\). Expanding, \((2x - 3)^2 = 4x^2 - 12x + 9\). Then \(gg(2x) = (4x^2 - 12x + 6)^2\). This can be expressed as \((2x - 3)^4 - 6(2x - 3)^2 + 9\).

(b)(ii) Expanding \((2x - 3)^4 - 6(2x - 3)^2 + 9\), we get \(16x^4 - 96x^3 + 216x^2 - 216x + 81 - 24x^2 + 72x - 54 + 9\). Simplifying, \(16x^4 - 96x^3 + 192x^2 - 144x + 36\).