8. Given $\angle A, \angle B, \angle C$ are the three interior angles of $\triangle ABC$, and the vector $\boldsymbol{\alpha}=\left(\cos \frac{A-B}{2}, \sqrt{3} \sin \frac{A+B}{2}\right),|\boldsymbol{\alpha}|=$ $\sqrt{2}$. If when $\angle C$ is maximized, there exists a moving point $M$, such that $|\overrightarrow{M A}|$, $|\overrightarrow{A B}|$, $|\overrightarrow{M B}|$ form an arithmetic sequence, then the maximum value of $\frac{|\overrightarrow{M C}|}{|\overrightarrow{A B}|}$ is
8. $\frac{2 \sqrt{3}+\sqrt{2}}{4}$.

Notice that,
$$
\begin{array}{l}
|\boldsymbol{\alpha}|=\sqrt{2} \Leftrightarrow \cos ^{2} \frac{A-B}{2}+3 \sin ^{2} \frac{A+B}{2}=2 \\
\Leftrightarrow \cos (A-B)=3 \cos (A+B) \\
\Leftrightarrow 2 \sin A \cdot \sin B=\cos A \cdot \cos B \\
\Leftrightarrow \tan A \cdot \tan B=\frac{1}{2} .
\end{array}
$$

Thus, $\tan C=\frac{\tan A+\tan B}{\tan A \cdot \tan B-1}$
$$
\begin{array}{l}
=-2(\tan A+\tan B) \\
\leqslant-4 \sqrt{\tan A \cdot \tan B}=-2 \sqrt{2},
\end{array}
$$

with equality holding if and only if $\tan A=\tan B=\frac{\sqrt{2}}{2}$. 
Let $|A B|=2 c$.
From $|\overrightarrow{M A}|+|\overrightarrow{M B}|=4 c$, we know that $M$ is a moving point on the ellipse
$$
\frac{x^{2}}{4 c^{2}}+\frac{y^{2}}{3 c^{2}}=1
$$

Thus, $C\left(0, \frac{\sqrt{2}}{2} c\right)$.
Let $M(x, y)(|y| \leqslant \sqrt{3} c)$. Then
$$
\begin{array}{l}
|M C|^{2}=x^{2}+\left(y-\frac{\sqrt{2}}{2} c\right)^{2} \\
=-\frac{1}{3} y^{2}-\sqrt{2} c y+\frac{9 c^{2}}{2} .
\end{array}
$$

When $y=-\sqrt{3} c$,
$$
\left(\frac{|\overrightarrow{M C}|}{|\overrightarrow{A B}|}\right)_{\max }=\frac{2 \sqrt{3}+\sqrt{2}}{4}
$$