1. Given non-zero vectors $\boldsymbol{a}, \boldsymbol{b}$ with an angle of $120^{\circ}$ between them. If vector $a-b$ is perpendicular to $a+2 b$, then $\left|\frac{2 a-b}{2 a+b}\right|=$ $\qquad$
- 1. $\frac{\sqrt{10+\sqrt{33}}}{3}$.

Let $|\boldsymbol{a}|=x,|\boldsymbol{b}|=y$.
According to the problem, we have $(4 x-y)^{2}=33 y^{2}$.
Also, $\left|\frac{2 a-b}{2 a+b}\right|=\sqrt{\frac{(2 a-b)^{2}}{(2 a+b)^{2}}}$
$$
=\sqrt{\frac{(4 x+y)^{2}+3 y^{2}}{(4 x-y)^{2}+3 y^{2}}}=\frac{\sqrt{10+\sqrt{33}}}{3} .
$$