7. If the parabola $y=x^{2}-a x+3(a \in \mathbf{R})$ is always above the line $y=\frac{9}{4}$ over the interval $[a, a+1]$, then the range of values for $a$ is $\qquad$.
7. Let $f(x)=y-\frac{9}{4}=x^{2}-a x+\frac{3}{4}=\left(x-\frac{a}{2}\right)^{2}+\frac{3-a^{2}}{4}$, the original problem is transformed into finding the range of $a$ such that the minimum value of $f(x)$ on $[a, a+1]$ is always positive.

Since $f(x)$ is monotonically decreasing on $\left(-\infty, \frac{a}{2}\right)$ and monotonically increasing on $\left(\frac{a}{2},+\infty\right)$, we have: (1) When $\frac{a}{2} \leqslant a$, i.e., $a \geqslant 0$, the minimum value of $f(x)$ on $[a, a+1]$ is $f(a)=\frac{3}{4}>0$, so we have $a \geqslant 0$; (2) When $a<0$, the minimum value of $f(x)$ on $[a, a+1]$ is $f\left(\frac{a}{2}\right)=\frac{3-a^{2}}{4}>0$, which implies $-\sqrt{3}<a<\sqrt{3}$. Since $a<0$, we have $-\sqrt{3}<a<0$. Additionally, from $f(a+1)=\left(a+1-\frac{a}{2}\right)^{2}+\frac{3-a^{2}}{4}>0$, we get $a>-\frac{7}{4}$. In this case, there is no solution for $a$.
In summary, the range of $a$ is $(-\sqrt{3},+\infty)$, fill in $(-\sqrt{3},+\infty)$.