1. Let $a=\operatorname{tg} x, b=\operatorname{tg} \frac{y}{2}$. Then $a+b=\frac{4}{\sqrt{3}}$ and $\frac{1}{a}+\frac{1}{b}=\frac{4}{\sqrt{3}}$. From the second equation, it follows that $a b=1$. By Vieta's theorem, $a$ and $b$ satisfy the equation $t^{2}-\frac{4}{\sqrt{3}} t+1=0$, whose roots are $\sqrt{3}$ and $\frac{1}{\sqrt{3}}$. Our system is equivalent to the combination of two systems: 1) $\left\{\begin{array}{l}\operatorname{tg} x=\sqrt{3}, \\ \operatorname{tg} \frac{y}{2}=\frac{1}{\sqrt{3}} ;\end{array}\right.$ or 2) $\left\{\begin{array}{l}\operatorname{tg} x=\frac{1}{\sqrt{3}} \\ \operatorname{tg} \frac{y}{2}=\sqrt{3}\end{array}\right.$
Answer: $x=\frac{\pi}{3}+\pi n, y=\frac{\pi}{3}+2 \pi k$ or $x=\frac{\pi}{6}+\pi n, y=\frac{2 \pi}{3}+2 \pi k$. Answer to option 1-2: $x=\frac{\pi}{3}+2 \pi k, y=\frac{\pi}{3}+\pi n$ or $x=\frac{2 \pi}{3}+2 \pi k, y=\frac{\pi}{6}+\pi n$.