(a) To express \(x^2 + 4x + 2\) in the form \((x+a)^2 + b\), complete the square:

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

Thus, \(a = 2\) and \(b = -2\).

(b)(i) To find \(f^{-1}(x)\), start with \(y = (x+2)^2 - 2\).

Rearrange to solve for \(x\):

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

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

Since \(x \leq -2\), choose the negative root:

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

(b)(ii) To find \((gf)^{-1}(x)\), first find \(gf(x)\):

\(gf(x) = g(f(x)) = g((x+2)^2 - 2) = -(x+2)^2 - 2 - 4 = -(x+2)^2 - 6\).

Now, solve \(y = -(x+2)^2 - 6\) for \(x\):

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

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

\(x+2 = \pm \sqrt{-y-6}\)

Since \(x \leq -2\), choose the negative root:

\((gf)^{-1}(x) = -\sqrt{-x-2} - 2\).