First, find the inverse of \(f\). Given \(f(x) = 3x + 2\), set \(y = 3x + 2\).

Solve for \(x\):

\(y - 2 = 3x\)

\(x = \frac{y - 2}{3}\)

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

Next, find \(gf(x)\):

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

Substitute into \(g\):

\(g(3x + 2) = 4(3x + 2) - 12\)

\(= 12x + 8 - 12\)

\(= 12x - 4\)

Equate \(f^{-1}(x)\) and \(gf(x)\):

\(\frac{x - 2}{3} = 12x - 4\)

Multiply through by 3 to clear the fraction:

\(x - 2 = 36x - 12\)

Rearrange and solve for \(x\):

\(x - 36x = -12 + 2\)

\(-35x = -10\)

\(x = \frac{2}{7}\)