To find the maximum possible value of \(a\), we need to ensure that the function \(f(x) = \frac{1}{3}(2x - 1)^{\frac{3}{2}} - 2x\) is decreasing for \(\frac{1}{2} < x < a\).

The derivative \(f'(x)\) is given by:

\(f'(x) = \left( (2x-1)^{\frac{1}{2}} \right) \times \left( \frac{1}{3} \times 2x \times \frac{3}{2} \right) \times (-2)\)

\(= (2x-1)^{\frac{1}{2}} \times (-2)\)

For \(f(x)\) to be decreasing, \(f'(x) \leq 0\).

\((2x-1)^{\frac{1}{2}} - 2 \leq 0\)

\(2x - 1 \leq 4\)

\(2x - 1 < 4\)

Solving \(2x - 1 < 4\), we get:

\(2x < 5\)

\(x < \frac{5}{2}\)

Thus, the maximum possible value of \(a\) is \(\frac{5}{2}\).