3. In $\triangle A B C$, $A B$ is the longest side, $\sin A \sin B=$ $\frac{2-\sqrt{3}}{4}$. Then the maximum value of $\cos A \cos B$ is $\qquad$ .
3. $\frac{2-\sqrt{3}}{4}$.

Given that $AB$ is the longest side, $A$ and $B$ are both acute angles, so we have
$$
\cos A=\sqrt{1-\sin ^{2} A}, \cos B=\sqrt{1-\sin ^{2} B} .
$$

Thus, $\cos A \cos B=\sqrt{1-\left(\sin ^{2} A+\sin ^{2} B\right)+\sin ^{2} A \sin ^{2} B}$
$$
\begin{array}{l}
=\sqrt{1-2 \sin A \sin B-(\sin A-\sin B)^{2}+\sin ^{2} A \sin ^{2} B} \\
=\sqrt{(1-\sin A \sin B)^{2}-(\sin A-\sin B)^{2}} \\
\leqslant 1-\sin A \sin B \\
=1-\frac{2-\sqrt{3}}{4}=\frac{2+\sqrt{3}}{4} . \\
\text { When } \sin A=\sin B=\sqrt{\frac{2-\sqrt{3}}{4}}=\frac{\sqrt{6}-\sqrt{2}}{4}
\end{array}
$$

$\cos A \cos B$ has the maximum value $\frac{2+\sqrt{3}}{4}$.