2. Let $0 \leqslant \alpha, \beta < 2 \pi, \alpha, \beta \neq \frac{\pi}{3}, \frac{5 \pi}{3}$, and
$$
\frac{2 \sin \alpha - \sqrt{3}}{2 \cos \alpha - 1} + \frac{2 \sin \beta - \sqrt{3}}{2 \cos \beta - 1} = 0 \text{. }
$$

Then $\cot \frac{\alpha + \beta}{2} = $ $\qquad$
2. $-\frac{\sqrt{3}}{3}$.

Notice that,
$$
\frac{2 \sin \alpha-\sqrt{3}}{2 \cos \alpha-1}+\frac{2 \sin \beta-\sqrt{3}}{2 \cos \beta-1}=0
$$

has a geometric meaning that the slopes of the lines connecting point $P\left(1, \frac{3}{2}\right)$ on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ to points $M(2 \cos \alpha, \sqrt{3} \sin \alpha)$ and $N(2 \cos \beta, \sqrt{3} \sin \beta)$ on the ellipse are opposite numbers.
The slope of the line $M N$ is
$k=\frac{\sqrt{3}(\sin \alpha-\sin \beta)}{2(\cos \alpha-\cos \beta)}=-\frac{\sqrt{3}}{2} \cot \frac{\alpha+\beta}{2}$.
Thus, $\cot \frac{\alpha+\beta}{2}=-\frac{2 \sqrt{3}}{3} k$ (where $k$ is a constant).
This constant $k$ is the slope of the tangent line to the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ at point $P\left(1,-\frac{3}{2}\right)$, which is $\frac{1}{2}$. Therefore,
$$
\cot \frac{\alpha+\beta}{2}=-\frac{2 \sqrt{3}}{3} \times \frac{1}{2}=-\frac{\sqrt{3}}{3} .
$$