3-ча 3. Find a four-digit number that, when divided by 131, gives a remainder of 112, and when divided by 132, gives a remainder of 98.
Solve 3. Answer: 1946. Let $N$ be the desired number. By the condition, $N=131 k+112=132 l+98$, where $k$ and $l$ are natural numbers. Moreover, $N<10000$, so $l=\frac{N-98}{132}<\frac{10000-98}{132} \leqslant 75$. Further, $131 k+112=$ $132 l+98$, so $131(k-l)=l-14$. Therefore, if $k \neq l$, then $|l-14| \geqslant 131$. But $l \leqslant 75$, so $k=l$ and $l-14=0$. Thus, $N=131 \cdot 14+112=132 \cdot 14+98=1946$.

Solve 4. Answer: $x_{1}=-x_{8}=1, x_{2}=-x_{2}=2, x_{3}=-x_{6}=3, x_{4}=-x_{5}=4$.

Adding all the equations, we get $3\left(x_{1}+x_{2}+\cdots+x_{8}\right)=0$. Then add the first, fourth, and seventh equations. The result is $2 x_{1}+x_{2}+x_{3}+\cdots+x_{8}=1$, so $x_{1}=1$. The other unknowns are found similarly.