7. $\frac{5}{2} \sqrt{3}$

Let the projection of $\mathrm{A}$ onto the lower base be $C$, and $CD$ be a diameter of the lower base. The midpoints of $AB, CB, OO_{1}$ are $L, M, N$ respectively. Then $LN$ is the distance from $AB$ to $OO_{1}$. By the Pythagorean theorem, $BC = \sqrt{AB^2 - AC^2} = 5$. And $BD = \sqrt{CD^2 - BC^2} = 5 \sqrt{3}$. Since $LV = MO_{1} = \frac{1}{2} BD$, we get $LN = \frac{5}{2} \sqrt{3}$.
7. Given the height of the cylinder $O O_{1}=12$, the base radius $r=5$, there is 1 point $A$ and 1 point $B$ on the circumference of the upper and lower base circles, respectively, and $A B=13$. Then, the distance between the axis of the cylinder $O O_{1}$ and $A B$ is