1. All integer pairs $(x, y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
y-\left|x^{2}-2 x\right|+\frac{1}{2}>0, \\
y+|x-1|<2
\end{array}\right.
$$

are $\qquad$
1. $(0,0),(1,1),(2,0)$.

From $y-\left|x^{2}-2 x\right|+\frac{1}{2}>0$ we get
$$
y>\left|x^{2}-2 x\right|-\frac{1}{2} \geqslant-\frac{1}{2} \text {; }
$$

From $y+|x-1|<2$ we get
$$
y<2-|x-1| \leqslant 2 \text {. }
$$

Therefore, $-\frac{1}{2}<y<2$.
So $y=0$ or 1.
When $y=0$, we have $\left\{\begin{array}{l}\left|x^{2}-2 x\right|<\frac{1}{2}, \\ |x-1|<2 .\end{array}\right.$
Solving this gives the integer $x=0$ or $x=2$.
When $y=1$, we have $\left\{\begin{array}{l}\left|x^{2}-2 x\right|<\frac{3}{2} \\ |x-1|<1 .\end{array}\right.$
Solving this gives the integer $x=1$.
Therefore, the integer solutions to the original inequality are $(0,0),(2,0),(1,1)$.