(i) To find \(fg(x)\), substitute \(g(x) = x^2 - 2\) into \(f(x) = 2x + 1\):

\(fg(x) = f(g(x)) = f(x^2 - 2) = 2(x^2 - 2) + 1 = 2x^2 - 4 + 1 = 2x^2 - 3\).

To find \(gf(x)\), substitute \(f(x) = 2x + 1\) into \(g(x) = x^2 - 2\):

\(gf(x) = g(f(x)) = g(2x + 1) = (2x + 1)^2 - 2 = 4x^2 + 4x + 1 - 2 = 4x^2 + 4x - 1\).

(ii) Set \(fg(a) = gf(a)\):

\(2a^2 - 3 = 4a^2 + 4a - 1\)

\(2a^2 - 3 = 4a^2 + 4a - 1 \Rightarrow 2a^2 + 4a + 2 = 0\)

\((a + 1)^2 = 0 \Rightarrow a = -1\).

(iii) Solve \(g(b) = b\):

\(b^2 - 2 = b \Rightarrow b^2 - b - 2 = 0\)

\((b + 1)(b - 2) = 0 \Rightarrow b = 2\) (also allow \(b = -1\)).

(iv) Find \(f^{-1}(x)\):

\(y = 2x + 1 \Rightarrow x = \frac{1}{2}(y - 1) \Rightarrow f^{-1}(x) = \frac{1}{2}(x - 1)\).

Substitute into \(g(x)\):

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

(v) Solve \(h(x) = x^2 - 2\) for \(x \leq 0\):

\(y = x^2 - 2 \Rightarrow x^2 = y + 2 \Rightarrow x = -\sqrt{y + 2} \Rightarrow h^{-1}(x) = -\sqrt{x + 2}\).