(a) To find \(a\), use \(f(7) = \frac{5}{2}\):

\(1 + \frac{2a}{7-a} = \frac{5}{2}\)

\(\frac{2a}{7-a} = \frac{3}{2}\)

\(4a = 3(7-a)\)

\(4a = 21 - 3a\)

\(7a = 21\)

\(a = 3\)

To find \(b\), use \(gf(5) = 4\):

\(f(5) = 1 + \frac{6}{5-3} = 4\)

\(g(4) = b(4) - 2 = 4\)

\(4b - 2 = 4\)

\(4b = 6\)

\(b = \frac{3}{2}\)

(b) The domain of \(f^{-1}\) is \(x > 1\) because \(f(x) = 1 + \frac{6}{x-3}\) implies \(x > 3\), and \(f(x) > 1\).

(c) To find \(f^{-1}(x)\):

Start with \(y = 1 + \frac{6}{x-3}\)

\(y - 1 = \frac{6}{x-3}\)

\((y-1)(x-3) = 6\)

\(yx - 3y = x - 3\)

\(yx - x = 3y - 3\)

\(x(y-1) = 3(y-1)\)

\(x = 3 + \frac{6}{y-1}\)

Thus, \(f^{-1}(x) = 3 + \frac{6}{x-1}\).