3. This system is equivalent to the following system:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ | \boldsymbol { y } - 3 x | \geq | 2 \boldsymbol { y } + \boldsymbol { x } | } \\
{ - 1 \leq y - 3 \leq 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(2 \boldsymbol{y}+\boldsymbol{x})^{2}-(\boldsymbol{y}-3 x)^{2} \leq 0 \\
2 \leq y \leq 4
\end{array}\right.\right. \\
& \left\{\begin{array} { l } 
{ ( 2 \boldsymbol { y } + \boldsymbol { y } - \boldsymbol { y } + 3 x ) ( 2 \boldsymbol { y } + \boldsymbol { x } + \boldsymbol { y } - 3 x ) \leq 0 } \\
{ 2 \leq y \leq 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(\boldsymbol{y}+4 x)(3 \boldsymbol{y}-2 x) \leq 0 \\
2 \leq \boldsymbol{y} \leq 4
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
(\boldsymbol{y}+4 x)\left(\boldsymbol{y}-\frac{2}{3} x\right) \leq 0 \\
2 \leq y \leq 4
\end{array}\right.
\end{aligned}
$$

We construct the lines $y=-4 x$ and $y=\frac{2}{3} \boldsymbol{x}$, and then the region, which will be a trapezoid with vertices $A(3 ; 2), B(-1 / 2 ; 2), C(-1 ; 4), D(6 ; 4)$. Its area is $S=\frac{1}{2} \cdot 2(7+3.5)=10.5$. Answer: $\{10.5\}$

4. Note that $\left(y^{2}\right)^{2}<(x+1)^{2}=y^{4}+y^{2}+1 \leq\left(y^{2}+1\right)^{2}$. From this, $y=0$ and $\mathrm{x}=0$ or $\mathrm{x}=-2$.
Answer: $(x ; y)=(0 ; 0)$ or $(-2 ; 0)$.