(i) To find \(gf(x)\), substitute \(g(x)\) into \(f\):

\(gf(x) = f(g(x)) = f(3x + 2) = 2(3x + 2)^2 + 3\).

Expanding, \((3x + 2)^2 = 9x^2 + 12x + 4\), so:

\(gf(x) = 2(9x^2 + 12x + 4) + 3 = 18x^2 + 24x + 8 + 3 = 18x^2 + 24x + 11\).

Thus, \(gf(x) = 6x^2 + 11\).

For \(fg(x)\), substitute \(f(x)\) into \(g\):

\(fg(x) = g(f(x)) = g(2x^2 + 3) = 3(2x^2 + 3) + 2 = 6x^2 + 9 + 2 = 6x^2 + 11\).

(ii) To find \((fg)^{-1}(x)\), start with \(y = fg(x) = 2(3x + 2)^2 + 3\).

Subtract 3: \(y - 3 = 2(3x + 2)^2\).

Divide by 2: \(\frac{y - 3}{2} = (3x + 2)^2\).

Take the square root: \(3x + 2 = \pm \sqrt{\frac{y - 3}{2}}\).

Subtract 2: \(3x = \pm \sqrt{\frac{y - 3}{2}} - 2\).

Divide by 3: \(x = \frac{1}{3}(\pm \sqrt{\frac{y - 3}{2}} - 2)\).

Thus, \((fg)^{-1}(x) = \frac{1}{3}\sqrt{\frac{x-3}{2}} - \frac{2}{3}\).

The domain is \(x \geq 11\) since \(fg(x)\) is defined for \(x > 0\).

(iii) Solve \(gf(2x) = fg(x)\):

\(gf(2x) = 6(2x)^2 + 11 = 24x^2 + 11\).

\(fg(x) = 6x^2 + 11\).

Equate: \(24x^2 + 11 = 6x^2 + 11\).

Simplify: \(18x^2 = 0\).

Thus, \(x = 0\).

Check for other solutions: \(6x^2 - 24x = 0\).

Factor: \(6x(x - 4) = 0\).

Solutions: \(x = 0, 4\).