(a) We have \(fg(x) = f(g(x)) = (g(x))^2 - 1\). Given \(fg(x) = 3\), we set \((g(x))^2 - 1 = 3\), leading to \((g(x))^2 = 4\).

Since \(g(x) = \frac{1}{2x+1}\), we have \(\left(\frac{1}{2x+1}\right)^2 = 4\).

This implies \(\frac{1}{(2x+1)^2} = 4\), leading to \((2x+1)^2 = \frac{1}{4}\).

Solving \(2x+1 = \pm \frac{1}{2}\), we get \(2x+1 = \frac{1}{2}\) or \(2x+1 = -\frac{1}{2}\).

For \(2x+1 = -\frac{1}{2}\), solving gives \(x = -\frac{3}{4}\).

For \(2x+1 = \frac{1}{2}\), solving gives \(x = -\frac{1}{4}\), but this does not satisfy \(x < -\frac{1}{2}\).

Thus, \(x = -\frac{3}{4}\) only.

(b) Let \(y = fg(x) = (g(x))^2 - 1\).

Then \(y = \frac{1}{(2x+1)^2} - 1\), leading to \(\frac{1}{(2x+1)^2} = y + 1\).

Thus, \((2x+1)^2 = \frac{1}{y+1}\).

Taking square roots, \(2x+1 = \pm \frac{1}{\sqrt{y+1}}\).

Solving for \(x\), we get \(x = -\frac{1}{2\sqrt{y+1}} - \frac{1}{2}\).