1. Given complex numbers $z_{1}, z_{2}$ satisfy $\left|z_{1}\right|=1,\left|z_{2}\right|=2,3 z_{1}-$ $z_{2}=2+\sqrt{3} \mathrm{i}$. Then $2 z_{1}+z_{2}=$ $\qquad$ .
II. $1.3-\sqrt{3} \mathrm{i}$ or $-\frac{9}{7}+\frac{13 \sqrt{3}}{7} \mathrm{i}$.
From $3 z_{1}-z_{2}=2+\sqrt{3} \mathrm{i}$, we get
$\left|3-\frac{z_{2}}{z_{1}}\right|=\frac{|2+\sqrt{3} i|}{\left|z_{1}\right|}=\sqrt{7}$.
Also, $\left|\frac{z_{2}}{z_{1}}\right|=\frac{\left|z_{2}\right|}{\left|z_{1}\right|}=2$, so $\frac{z_{2}}{z_{1}}=1 \pm \sqrt{3} \mathrm{i}$.
Thus, $z_{1}=\frac{2+\sqrt{3} \mathrm{i}}{3-\frac{z_{2}}{z_{1}}}=1$ or $\frac{1+4 \sqrt{3} \mathrm{i}}{7}$.
Therefore, $2 z_{1}+z_{2}=5 z_{1}-\left(3 z_{1}-z_{2}\right)=3-\sqrt{3} \mathrm{i}$ or $-\frac{9}{7}+\frac{13 \sqrt{3}}{7} \mathrm{i}$.