3. In the tetrahedron $P-ABC$, $PA=PB=a$, $PC=AB=BC=CA=b$. And $a<b$. Then the range of $\frac{a}{b}$ is

保留了源文本的换行和格式。
3. $(\sqrt{2-\sqrt{3}}, 1)$.

Regardless of what positive number $b$ takes, $\triangle A B C$ always exists, and its height is $\frac{\sqrt{3}}{2} b$. For $\triangle P A B$ to exist, it should have $bP C$.
$$
\text{Rearranging and substituting, we get } \sqrt{a^{2}-\frac{b^{2}}{4}}>b-\frac{\sqrt{3}}{2} b,
$$
Squaring both sides, we get $a^{2}>(2-\sqrt{3}) b^{2}$,
Thus, $\frac{a}{b}>\sqrt{2-\sqrt{3}}$.
Therefore, the range of $\frac{a}{b}$ is $(\sqrt{2-\sqrt{3}}, 1)$.
Note: For $a, b$ that satisfy this range, we can first construct $\triangle P C D$, then determine $A$ and $B$. Further consideration of a tetrahedron with edge lengths taking two values $a, b$ is quite interesting.