$\xi = \{(x,y): x \text{ 和 } y \text{ 是整数}, -2 \le x < 1 \text{ 且 } -1 \le y \le 1\}$ $P = \{(x,y): x^2 + y^2 < 2\}$ $Q = \{(x,y): xy < 0\}$

1. Count the possible ordered pairs in the universal set

Since $x$ and $y$ are integers,

$x=-2,-1,0$ gives $3$ possible values.

$y=-1,0,1$ gives $3$ possible values.

Hence,
$n(\xi)=3\times 3=9.$

1. Identify the possible ordered pairs from the universal set and test the condition for $P$

From $\xi$, $x\in\{-2,-1,0\}$ and $y\in\{-1,0,1\}$.

For $P$, we need
$x^2+y^2<2.$
Testing these possible ordered pairs, the pairs that satisfy this are
$(-1,0),\ (0,-1),\ (0,0),\ (0,1).$
Hence,
$P=\{(-1,0),(0,-1),(0,0),(0,1)\}.$

1. Identify the ordered pairs in the universal set that satisfy $xy<0$

From $\xi$, $x\in\{-2,-1,0\}$ and $y\in\{-1,0,1\}$. Test $xy<0$:

$xy<0$ means $x$ and $y$ have opposite signs.

Since $x$ can be negative or zero only, the valid pairs are:

$(-2,1)$ and $(-1,1)$.