2. Find the ordered integer pairs $(x, y)$ that satisfy $(|x|-2)^{2}+(|y|-2)^{2}<5$.

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2. Clearly, $(|x|-2)^{2}$ and $(|y|-2)^{2}$ are both perfect squares.

Since the sum of these two perfect square terms is less than 5, each term must be no greater than $2^{2}=4$.

If $(|x|-2)^{2}=0$, then $|x|=2$. Thus, $x= \pm 2$. Therefore, $x$ has two possible values.

If $(|x|-2)^{2}=1$, then $|x|=1$ or $|x|=3$. Thus, $x$ $= \pm 1$ or $x= \pm 3$. Therefore, $x$ has four possible values.

If $(|x|-2)^{2}=4$, then $|x|=0$ or $|x|=4$. Thus, $x$ $=0$ or $x= \pm 4$. Therefore, $x$ has three possible values.

If $(|x|-2)^{2}=0$ and $(|y|-2)^{2}=0, 1$ or 2, the total number of possible ordered integer pairs is $2 \times(2+4+3)=18$.

If $(|x|-2)^{2}=1$ and $(|y|-2)^{2}=0, 1$, the total number of possible ordered integer pairs is $4 \times(2+4)=24$.

If $(|x|-2)^{2}=4$ and $(|y|-2)^{2}=0$, the total number of possible ordered integer pairs is $3 \times 2=6$.

In summary, the total number of ordered integer pairs is $18+24+6=48$.