8. Given the set $M=\left\{x \mid x^{2}-9>0\right\}, N=\left\{x \in \mathbf{Z} \mid x^{2}-8 x+a<0\right\}$. If the number of subsets of the set $M \cap N$ is 4, then the range of the real number $a$ is $\qquad$ .
$M=(-\infty,-3) \bigcup(3,+\infty)$, and the axis of symmetry of $f(x)=x^{2}-8 x+a$ is $x=4$, so we have $M \cap N=\{4,5\}$. Thus, $\left\{\begin{array}{l}f(5)=25-40+a<0, \\ f(6)=36-48+a \geqslant 0\end{array} \Rightarrow 12 \leqslant a<15\right.$.
Therefore, the range of the real number $a$ is $[12,15)$.