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