3. As shown in Figure 4, the parabola $y=a x^{2}+b x-3$ intersects the $x$-axis at points $A$ and $B$, and intersects the $y$-axis at point $C$, and $O B=O C=3 O A$. The line $y=-\frac{1}{3} x+1$ intersects the $y$-axis at point $D$, and $E$ is the vertex of the parabola. If $\angle D B C=\alpha, \angle C B E=\beta$, then $\alpha-\beta=$ . $\qquad$
3.45.

It is known that $D(0,1), C(0,-3), A(-1,0)$, $B(3,0)$.
Then $y=a(x+1)(x-3) \Rightarrow a=1$.
Thus, $y=x^{2}-2x-3$, and its vertex is $E(1,-4)$.
Therefore, $BC=3\sqrt{2}, CE=\sqrt{2}, BE=2\sqrt{5}$.
Since $BC^{2}+CE^{2}=BE^{2}$, $\triangle BCE$ is a right triangle.
Thus, $\tan \beta=\frac{CE}{CB}=\frac{1}{3}$.
Also, $\tan \angle DBO=\frac{OD}{OB}=\frac{1}{3}$, so
$\angle DBO=\beta$.
Therefore, $\alpha-\beta=\alpha-\angle DBO=\angle OBC=45^{\circ}$.