To determine where the function \(\(f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 6x + 7\)\) is decreasing and increasing, we need to find its critical points and analyze the signs of the derivative.

First, we calculate the derivative of the function:

\(f'(x) = \frac{d}{dx} \left( \frac{x^3}{3} - \frac{5x^2}{2} + 6x + 7 \right) = x^2 - 5x + 6\)

The behavior of the function changes at the points where the derivative is zero (critical points) or undefined. This derivative is defined for all \(x\), so we only need to find where \(f'(x) = 0\):

\(x^2 - 5x + 6 = 0\)

Factoring the quadratic gives:

\((x - 2)(x - 3) = 0\)

Therefore, the critical points are \(x = 2\) and \(x = 3\).

To determine the intervals of increase and decrease, we test intervals around the critical points \(x = 2\) and \(x = 3\):

1. For \(x \lt 2\), choose \(x = 1\):
2. For \(2 \lt x \lt 3\), choose \(x = 2.5\):
3. For \(x \gt 3\), choose \(x = 4\):

From these calculations, we observe that:

• The function decreases on the interval \((2, 3)\).
• The function increases on the intervals \((-∞, 2)\) and \((3, ∞)\).

The set \(T\), where the function increases, corresponds to \(\{x \lt 2\} \cup \{x \gt 3\}\).

Thus, the correct answer is: \(\{x \lt 2\} \cup \{x \gt 3\}\).