3. Given $O$ is the circumcenter of $\triangle A B C$, $|A B|=2,|A C|=1, \angle B A C=\frac{2}{3} \pi$, let $\overrightarrow{A B}=\boldsymbol{a}, \overrightarrow{A C}=\boldsymbol{b}$, if $\overrightarrow{A O}=\lambda_{1} a+\lambda_{2} \boldsymbol{b}$, then $\lambda_{1}+\lambda_{2}=$ $\qquad$
3. $\frac{13}{6}$, Detailed Explanation: Establish a Cartesian coordinate system, $A(0,0), B(2,0), C\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$. Clearly, the midpoint of $A C$ is $M\left(-\frac{1}{4}, \frac{\sqrt{3}}{4}\right)$, and we can set $O(1, y)$. Since $O M \perp A C$, we have $\overrightarrow{O M} \cdot \overrightarrow{A C}=0$, which is $-\frac{5}{4} \times\left(-\frac{1}{2}\right)+\left(\frac{\sqrt{3}}{4}-y\right) \times \frac{\sqrt{3}}{2}=0$. Solving this, we get $y=\frac{2 \sqrt{3}}{3}$, so $\overrightarrow{A O}=\left(1, \frac{2 \sqrt{3}}{3}\right)$. From $\overrightarrow{A O}=\lambda_{1} a+\lambda_{2} b$, we have $1=2 \lambda_{1}+\left(-\frac{1}{2}\right) \lambda_{2}$, and $\frac{2 \sqrt{3}}{3}=\frac{\sqrt{3}}{2} \lambda_{2}$. Solving these, we get $\lambda_{2}=\frac{4}{3}, \lambda_{1}=\frac{5}{6}$, hence $\lambda_{1}+\lambda_{2}=\frac{13}{6}$.