(a) To find \(b\), use \(gg(2) = 10\):

\(g(x) = 3x + b\), so \(g(2) = 3(2) + b = 6 + b\).

Then \(gg(2) = g(g(2)) = g(6 + b) = 3(6 + b) + b = 18 + 3b + b = 18 + 4b\).

Set \(18 + 4b = 10\), solving gives \(b = -2\).

To find \(a\), use \(f^{-1}(2) = 14\):

\(f(x) = \frac{1}{2}x - a\), so \(f^{-1}(x) = 2(x + a)\).

Set \(2(x + a) = 14\), solving gives \(a = 5\).

(b) Find \(gf(x)\):

\(gf(x) = g(f(x)) = g\left(\frac{1}{2}x - 5\right)\).

\(g(x) = 3x + b\), so \(g\left(\frac{1}{2}x - 5\right) = 3\left(\frac{1}{2}x - 5\right) - 2\).

\(= \frac{3}{2}x - 15 - 2 = \frac{3}{2}x - 17\).