(a) To find \(ff(5)\), we first calculate \(f(5)\):

\(f(5) = \frac{5+3}{5-1} = \frac{8}{4} = 2\).

Now, calculate \(f(f(5)) = f(2)\):

\(f(2) = \frac{2+3}{2-1} = \frac{5}{1} = 5\).

Thus, \(ff(5) = 5\).

(b) To find \(f^{-1}(x)\), set \(y = f(x) = \frac{x+3}{x-1}\).

Swap \(x\) and \(y\):

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

Cross-multiply to solve for \(y\):

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

\(xy - x = y + 3\).

\(xy - y = x + 3\).

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

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

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