3.3. $\left\{\begin{array}{l}2 y=4-x^{2}, \\ 2 x=4-y^{2} .\end{array}\right.$
3.3. Subtracting the second equation from the first, we get $2(y-x)=$ $=y^{2}-x^{2}=(y-x)(y+x)$. Therefore, either $y=x$ or $y+x=2$. If $y=x$, then $2x=4-x^{2}$, i.e., $x=-1 \pm \sqrt{5}$. If $y+x=2$, then $2(2-x)=4-x^{2}$, i.e., $x^{2}=2x$. As a result, we obtain four solutions: $(-1+\sqrt{5},-1+\sqrt{5}),(-1-\sqrt{5},-1-\sqrt{5}),(0,2),(2,0)$.