(i) To evaluate \(fg(2)\), first find \(g(2)\):

\(g(2) = 2(2 - 1)^3 + 8 = 2(1)^3 + 8 = 2 + 8 = 10\).

Then, find \(f(g(2)) = f(10)\):

\(f(10) = 3(10) - 4 = 30 - 4 = 26\).

Therefore, \(fg(2) = 26\).

(iv) To find \(f^{-1}(x)\), solve \(y = 3x - 4\) for \(x\):

\(y + 4 = 3x\)

\(x = \frac{y + 4}{3}\)

Thus, \(f^{-1}(x) = \frac{x + 4}{3}\).

To find \(g^{-1}(x)\), solve \(y = 2(x - 1)^3 + 8\) for \(x\):

\(y - 8 = 2(x - 1)^3\)

\(\frac{y - 8}{2} = (x - 1)^3\)

\(x - 1 = \sqrt[3]{\frac{y - 8}{2}}\)

\(x = \sqrt[3]{\frac{y - 8}{2}} + 1\)

Thus, \(g^{-1}(x) = \sqrt[3]{\frac{x - 8}{2}} + 1\).