8. Given $a=\frac{11 \cdots 1}{2009 \uparrow}, b=\frac{100 \cdots 05}{2008 \uparrow}$. Then $10^{2009}-\sqrt{a b+1}=$ $\qquad$
8. $\underbrace{66 \cdots 6}_{20091}$.

From the given, we have
$$
\begin{array}{l}
a=\frac{11 \cdots 1}{2009 \uparrow}=\sum_{i=0}^{2008} 10^{i}=\frac{10^{2009}-1}{9}, \\
b=\underset{2008 \uparrow}{100 \cdots 05}=10^{2009}+5 .
\end{array}
$$

Then $a b+1=\frac{10^{2009}-1}{9}\left(10^{2009}+5\right)+1$
$$
\begin{array}{l}
=\frac{\left(10^{2009}\right)^{2}+4 \times 10^{2009}+4}{9} \\
=\left(\frac{10^{2009}+2}{3}\right)^{2}=\left(\frac{10^{2009}-10+12}{3}\right)^{2} \\
=\left(\frac{99 \cdots 90}{3}+4\right)^{2}=\left(\frac{33 \cdots 34}{2008 \uparrow}\right)^{2} .
\end{array}
$$

Therefore, $10^{2009}-\sqrt{a b+1}$
$$
=10^{2009}-\underbrace{33 \cdots 3}_{2008 \uparrow} 4=\underbrace{66 \cdots}_{2009 \uparrow} .
$$