To find \(\int \frac{2}{\sqrt{5x - 6}} \, dx\), we use substitution. Let \(u = 5x - 6\), then \(du = 5 \, dx\) or \(dx = \frac{du}{5}\).

The integral becomes \(\int \frac{2}{\sqrt{u}} \cdot \frac{1}{5} \, du = \frac{2}{5} \int u^{-1/2} \, du\).

The integral of \(u^{-1/2}\) is \(2u^{1/2}\), so we have:

\(\frac{2}{5} \cdot 2u^{1/2} = \frac{4}{5} \sqrt{u}\).

Substituting back for \(u\), we get \(\frac{4}{5} \sqrt{5x - 6}\).

Now, evaluate \(\int_{2}^{3} \frac{2}{\sqrt{5x - 6}} \, dx\):

\(\left[ \frac{4}{5} \sqrt{5x - 6} \right]_{2}^{3} = \frac{4}{5} \left( \sqrt{9} - \sqrt{4} \right)\).

\(= \frac{4}{5} (3 - 2) = \frac{4}{5} \cdot 1 = \frac{4}{5} = 0.8\).