3. Given the point sets
$$
\begin{array}{l}
A=\left\{(x, y) \left\lvert\,(x-3)^{2}+(y-4)^{2} \leqslant\left(\frac{5}{2}\right)^{2}\right.\right\}, \\
B=\left\{(x, y) \left\lvert\,(x-4)^{2}+(y-5)^{2}>\left(\frac{5}{2}\right)^{2}\right.\right\} .
\end{array}
$$

Then the number of integer points (i.e., points with both coordinates as integers) in the point set $A \cap B$ is
3. 7.

As shown in the figure, circles $E$ and $F$ intersect at points $M$ and $N$. The entire figure is symmetric about the line connecting the centers $E F$. Among them, $A \cap B$ is a crescent shape $S$ near the origin $O$ on the lower left. The x-coordinate of the points in $S$ can be 1, 2, 3, 4, and the y-coordinate can be 2, 3, 4, 5. The axis of symmetry $E F$ passes through the crescent shape $S$, and it is found that it passes through only one integer point $C_{4}(2,3)$. The integer points in $S$ with an x-coordinate of 1 are 3:
$$
C_{1}(1,5), C_{2}(1,4), C_{3}(1,3).
$$

The axis-symmetric points of these three points are, in order,
$$
C_{5}(2,2), C_{0}(3,2), C_{7}(4,2).
$$

Therefore, there are a total of 7 integer points.