3. One focus of a hyperbola is $F(0,4)$, the length of the real axis is 6, and point $A(0,3)$ lies on this hyperbola. Then the maximum value of the eccentricity of this hyperbola is $\qquad$
3. $\frac{4}{3}$.

Let the center of the hyperbola be $O^{\prime}(x, y)$, and the other focus be $F^{\prime}(x_{0}, y_{0})$. Since $O^{\prime}$ is the midpoint of $F F^{\prime}$, then $x_{0}=2 x, y_{0}=2 y-4$.
Since $A(0,3)$ is on the hyperbola, then
$||A F^{\prime}|-| A F||=2 a=6$,

i.e., $\left|\sqrt{(2 x)^{2}+(2 y-7)^{2}}-1\right|=6$.
Thus, $\sqrt{(2 x)^{2}+(2 y-7)^{2}}=7$.
Therefore, $x^{2}+\left(y-\frac{7}{2}\right)^{2}=\frac{49}{4}$.
[Since $\left|F F^{\prime}\right|>6$, then $y \neq 7, 0 \leqslant y<7$.
The semi-focal distance $c: \left|O^{\prime} F\right|=\sqrt{x^{2}+(y-4)^{2}}$, then
$$
\begin{array}{l}
e=\frac{c}{a}=\frac{\sqrt{x^{2}+(y-4)^{2}}}{3} \\
=\frac{1}{3} \sqrt{\frac{49}{4}-\left(y-\frac{7}{2}\right)^{2}+(y-4)^{2}} \\
=\frac{1}{3} \sqrt{16-y} \quad(0 \leqslant y<7) .
\end{array}
$$

When $y=0$, $e$ reaches its maximum value.