3. Given the function $f(x)=\left|x^{2}-2\right|$. If $f(a)=$ $f(b)$, and $0<a<b$, then the range of values for $a b$ is
3. $(0,2)$.

From Figure 2, based on the graph of $y=f(x)$, the only values that satisfy $f(a)=f(b)$ and $0<a<b$ are
$$
\begin{array}{c}
0<a<\sqrt{2} \\
<b<2,
\end{array}
$$

and $f(a)=2-a^{2}, f(b)=b^{2}-2$.
Thus, $2-a^{2}=b^{2}-2$, which simplifies to $a^{2}+b^{2}=4$.
Therefore, $a$ and $b$ can be expressed as
$$
b=2 \cos \theta, a=2 \sin \theta\left(0<\theta<\frac{\pi}{4}\right) \text {. }
$$

Hence, $a b=4 \cos \theta \cdot \sin \theta=2 \sin 2 \theta$.
Since $0<2 \theta<\frac{\pi}{2}$, the range of $a b$ is $(0,2)$.