Given the equation \(y = a + \tan bx\), we know the curve intersects the \(y\)-axis at \((0, \sqrt{3})\). Substituting \(x = 0\) into the equation gives:

\(\sqrt{3} = a + \tan(0)\)

\(\sqrt{3} = a\)

Thus, \(a = \sqrt{3}\).

Next, the curve intersects the \(x\)-axis at \(\left(-\frac{1}{6}\pi, 0\right)\). Substituting \(y = 0\) and \(x = -\frac{1}{6}\pi\) into the equation gives:

\(0 = \sqrt{3} + \tan\left(-\frac{b}{6}\pi\right)\)

\(\tan\left(-\frac{b}{6}\pi\right) = -\sqrt{3}\)

The angle whose tangent is \(-\sqrt{3}\) is \(-\frac{\pi}{3}\), so:

\(-\frac{b}{6}\pi = -\frac{\pi}{3}\)

Solving for \(b\):

\(\frac{b}{6} = \frac{1}{3}\)

\(b = 2\)

Thus, \(b = 2\).