6. Given that $\alpha$ is an acute angle, vectors
$$
a=(\cos \alpha, \sin \alpha), b=(1,-1)
$$

satisfy $a \cdot b=\frac{2 \sqrt{2}}{3}$. Then $\sin \left(\alpha+\frac{5 \pi}{12}\right)=$ $\qquad$
6. $\frac{2+\sqrt{15}}{6}$.

From the problem, we know
$$
\cos \alpha - \sin \alpha = \frac{2 \sqrt{2}}{3} \Rightarrow \cos \left(\alpha + \frac{\pi}{4}\right) = \frac{2}{3} > 0 \text{. }
$$

Since $\alpha$ is an acute angle, we have
$$
\sin \left(\alpha + \frac{\pi}{4}\right) = \frac{\sqrt{5}}{3} \text{. }
$$

Therefore, $\sin \left(\alpha + \frac{5 \pi}{12}\right) = \sin \left[\left(\alpha + \frac{\pi}{4}\right) + \frac{\pi}{6}\right]$
$$
= \frac{\sqrt{5}}{3} \times \frac{\sqrt{3}}{2} + \frac{2}{3} \times \frac{1}{2} = \frac{2 + \sqrt{15}}{6} \text{. }
$$