1. Given non-zero vectors $\boldsymbol{a} 、 \boldsymbol{b}$ with an angle of $120^{\circ}$ between them. If vector $\boldsymbol{a}-\boldsymbol{b}$ is perpendicular to $\boldsymbol{a}+2 \boldsymbol{b}$, then $\left|\frac{2 a-b}{2 a+b}\right|=$
$-1 . \frac{\sqrt{10+\sqrt{33}}}{3}$.
Let $|\boldsymbol{a}|=x,|\boldsymbol{b}|=y$.
According to the problem, $(4 x-y)^{2}=33 y^{2}$.
Also, $\left|\frac{2 a-b}{2 a+b}\right|=\sqrt{\frac{(2 a-b)^{2}}{(2 a+b)^{2}}}$
$=\sqrt{\frac{(4 x+y)^{2}+3 y^{2}}{(4 x-y)^{2}+3 y^{2}}}=\frac{\sqrt{10+\sqrt{33}}}{3}$.