To determine when the function \(f(x) = x^3 - x^2 - 8x + 5\) is increasing, we need to find where its derivative is positive.

First, differentiate \(f(x)\):

\(f'(x) = 3x^2 - 2x - 8\).

Set \(f'(x) > 0\) to find where the function is increasing:

\(3x^2 - 2x - 8 > 0\).

Solving the inequality \(3x^2 - 2x - 8 = 0\) gives the critical points. Using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -2\), \(c = -8\).

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

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

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

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

\(x = 2\) or \(x = -\frac{4}{3}\).

The function is increasing in the interval \(x < -\frac{4}{3}\) and \(x > 2\).

Since \(x < a\), the largest possible value of \(a\) is \(-\frac{4}{3}\).