(a) To find \(fg(7)\), we first calculate \(g(7)\):

\(g(7) = \frac{4}{7+1} = \frac{4}{8} = \frac{1}{2}\).

Then, calculate \(f(g(7)) = f\left(\frac{1}{2}\right)\):

\(f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right) - 2 = 2 - 2 = 0\).

Thus, \(fg(7) = 0\).

(b) To find \(x\) for which \(f^{-1}(x) = g^{-1}(x)\), we first find the inverses:

\(f(x) = 4x - 2 \Rightarrow y = 4x - 2 \Rightarrow x = \frac{y+2}{4} \Rightarrow f^{-1}(x) = \frac{x+2}{4}\).

\(g(x) = \frac{4}{x+1} \Rightarrow y = \frac{4}{x+1} \Rightarrow y(x+1) = 4 \Rightarrow x = \frac{4}{y} - 1 \Rightarrow g^{-1}(x) = \frac{4}{x} - 1\).

Set \(f^{-1}(x) = g^{-1}(x)\):

\(\frac{x+2}{4} = \frac{4}{x} - 1\).

Multiply through by 4x to clear fractions:

\(x(x+2) = 16 - 4x\).

\(x^2 + 2x = 16 - 4x\).

\(x^2 + 6x - 16 = 0\).

Factor the quadratic:

\((x+8)(x-2) = 0\).

Thus, \(x = -8\) or \(x = 2\).