To find the stationary points, we first find the derivative of the curve:

\(\frac{dy}{dx} = 3x^2 + 2x - 8\).

Set \(\frac{dy}{dx} = 0\) to find the stationary points:

\(3x^2 + 2x - 8 = 0\).

Solving this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 2\), and \(c = -8\), we find:

\(x = \frac{-2 \pm \sqrt{2^2 - 4 \times 3 \times (-8)}}{2 \times 3}\).

\(x = \frac{-2 \pm \sqrt{4 + 96}}{6}\).

\(x = \frac{-2 \pm \sqrt{100}}{6}\).

\(x = \frac{-2 \pm 10}{6}\).

This gives the solutions \(x = \frac{8}{6} = \frac{4}{3}\) and \(x = \frac{-12}{6} = -2\).

For the curve to have no stationary points in the interval \(a < x < b\), the interval must not include these solutions. Therefore, the least possible value of \(a\) is \(-2\) and the greatest possible value of \(b\) is \(\frac{4}{3}\).