7. Let $[x]$ denote the greatest integer not greater than $x$, and the sets
$$
\begin{array}{l}
A=\left\{x \mid x^{2}-2[x]=3\right\}, \\
B=\left\{x \left\lvert\, \frac{1}{8}<2^{x}<8\right.\right\} .
\end{array}
$$

Then $A \cap B=$ $\qquad$
$=7 . \mid-1, \sqrt{7}\}$.
Since the solution to the inequality $\frac{1}{8}<2^{x}<8$ is $-3<x<3$, therefore, $B=(-3,3)$.
When $x \in A \cap B$, $\left\{\begin{array}{l}x^{2}-2[x]=3, \\ -3<x<3 .\end{array}\right.$
Thus, $[x]$ can only take the values $-3,-2,-1,0,1,2$.
If $[x] \leqslant-2$, then $x^{2}=3+2[x]<0$, which has no real solutions;

If $[x]=-1$, then $x^{2}=1$, solving gives $x=-1$; if $[x]=0$, then $x^{2}=3$, which has no valid solutions; if $[x]=1$, then $x^{2}=5$, which has no valid solutions; if $[x]=2$, then $x^{2}=7$, solving gives $x=\sqrt{7}$.
Therefore, $A \cap B=\{-1, \sqrt{7}\}$.