8. For any positive integer $n$, we have
$$
\begin{array}{l}
\frac{1}{n}\left[\sqrt{1-\left(\frac{1}{n}\right)^{2}}+\sqrt{1-\left(\frac{2}{n}\right)^{2}}+\cdots+\sqrt{1-\left(\frac{n-1}{n}\right)^{2}}\right] \\
<a<\frac{1}{n}\left[1+\sqrt{1-\left(\frac{1}{n}\right)^{2}}+\cdots+\sqrt{1-\left(\frac{n-1}{n}\right)^{2}}\right] .
\end{array}
$$

Then the real number $a=$
8. $\frac{\pi}{4}$.

Construct the function $y=\sqrt{1-x^{2}}(x \geqslant 0)$, its graph is the part of the unit circle in the first quadrant, as shown in Figure 4.

Let $A(1,0)$ and $B(0,1)$. Divide the line segment $O A$ into $n$ equal parts, and draw perpendiculars from each of these points and point $A$ to the $x$-axis, intersecting the unit circle at $n-1$ points. Then draw horizontal lines from point $B$ and the $n-1$ intersection points, forming $n-1$ "small rectangles" (one vertex on the circle) and $n$ "large rectangles" (one vertex outside the circle) within the circle.
It is easy to see that the sum of the areas of the $n-1$ small rectangles is
$$
S_{1}=\frac{1}{n}\left[\sqrt{1-\left(\frac{1}{n}\right)^{2}}+\sqrt{1-\left(\frac{2}{n}\right)^{2}}+\cdots+\sqrt{1-\left(\frac{n-1}{a}\right)^{2}}\right],
$$
and the sum of the areas of the $n$ large rectangles is
$$
S_{2}=\frac{1}{n}\left[1+\sqrt{1-\left(\frac{1}{n}\right)^{2}}+\cdots+\sqrt{1-\left(\frac{n-1}{n}\right)^{2}}\right].
$$

The area of the part of the unit circle in the first quadrant is $S=\frac{\pi}{4}$, clearly, $S_{1}<\frac{\pi}{4}<S_{2}$.
Furthermore, as $n \rightarrow+\infty$, $S_{1} \rightarrow \frac{\pi}{4}, S_{2} \rightarrow \frac{\pi}{4}$.
Therefore, for any positive integer $n$, the inequality $S_{1}<a<S_{2}$ holds if and only if $a=\frac{\pi}{4}$.