7 If the point with the abscissa $\frac{3}{2} a$ on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is farther from the right focus than from the left directrix, then the range of the acute angle between the two asymptotes of the hyperbola is . $\qquad$
(7) The left directrix of the hyperbola is given by $x=-\frac{a^{2}}{c}$, and the right directrix is given by $x=\frac{a^{2}}{c}$. For a point on the hyperbola with the x-coordinate $\frac{3}{2} a$, the distance to the right focus is $e \cdot\left|\frac{3}{2} a-\frac{a^{2}}{c}\right|$, and the distance to the left directrix is $\left|\frac{3}{2} a-\left(-\frac{a^{2}}{c}\right)\right|$. According to the condition,
$$
e \cdot\left|\frac{3}{2} a-\frac{a^{2}}{c}\right|>\left|\frac{3}{2} a-\left(-\frac{a^{2}}{c}\right)\right|,
$$

simplifying gives
$$
\left|\frac{3}{2} e-1\right|>\left|\frac{3}{2}+\frac{1}{e}\right| \text {. }
$$

Combining with $e>1$, we solve to get $e>2$. Therefore, $\frac{\sqrt{a^{2}+b^{2}}}{a}>2$, which means $\frac{b}{a}>\sqrt{3}$. This implies that the acute angle between the two asymptotes of the hyperbola is in the range $\left(0^{\circ}, 60^{\circ}\right)$.