(a) We are looking for integers $m$ such that
$$10^{m}-1=13 n+8 \text{ for some positive integer } n$$
The last relation is equivalent to $10^{m}-9 \equiv 0 \pmod{13}$.
We will prove that $\operatorname{ord}_{13}(10)=6$, that is, the smallest positive integer $s$ such that $10^{s} \equiv 1 \pmod{13}$ is $s=6$. Indeed, we have
$$10^{6}=(13-3)^{6} \equiv 3^{6}=(3^{3})^{2}=(26+1)^{2} \equiv 1 \quad (\bmod\ 13).$$
Moreover, $10^{2} \equiv 9 \pmod{13}$ and $10^{3} \equiv -1 \pmod{13}$, hence
$$\operatorname{ord}_{13}(10)=6.$$
From the previous property, it follows that for any integer $k \geq 0$, we have $10^{6k} \equiv 1 \pmod{13}$. Therefore,
$$10^{6k+2}=10^{6k} \cdot 10^{2} \equiv 9 \quad (\bmod\ 13),$$
hence the integers
$$10^{6k+2}-1=\underbrace{99\ldots 9}_{6k+2},\ k=0,1,\ldots,$$
belong to the sequence, and all integers in the sequence containing only digit $9$ are of this form.

(b) It is clear that we have to find the greatest integer $k$ such that $10^{6k+2}-1<2010^{2010}$, that is $10^{6k+2}-1<10^{6k+2}<2010^{2010}$. It follows $6k+2<2010 \log 2010$, hence
$$\begin{gathered} 3k+1<1005 \log 201 \cdot 10=1005+1005 \log 201 \\ =1005+2010+1005 \log \left(2+\frac{1}{100}\right) \\ =3015+1005 \log \left(2+\frac{1}{100}\right) \end{gathered}$$
hence
$$k=1004+\left[\frac{2}{3}+335 \log \left(2+\frac{1}{100}\right)\right]=1105.$$
The desired number is $1106$.