3. Given the quadratic function
$$
y=3 a x^{2}+2 b x-(a+b) \text {, }
$$

when $x=0$ and $x=1$, the value of $y$ is positive. Then, when $0<x<1$, the parabola intersects the $x$-axis at $\qquad$ points.
3. 2 .

From the given, we have $\left\{\begin{array}{l}-(a+b)>0, \\ 3 a+2 b-(a+b)>0 .\end{array}\right.$
Solving, we get $a>0$.
Let $x=\frac{1}{2}$, then
$$
y=\frac{3 a}{4}+b-(a+b)=-\frac{a}{4}<0 .
$$

By the Intermediate Value Theorem, $3 a x^{2}+2 b x-(a+b)=0$ has one root when $0<x<\frac{1}{2}$, and one root when $\frac{1}{2}<x<1$. Therefore, there are two intersection points.