2. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=a(a \neq 0$, and $a \neq 1)$, the sum of the first $n$ terms is $S_{n}$, and $S_{n}=\frac{a}{1-a}\left(1-a_{n}\right)$. Let $b_{n}=a_{n} \lg \left|a_{n}\right|\left(n \in \mathbf{N}_{+}\right)$. When $a=-\frac{\sqrt{7}}{3}$, does there exist a positive integer $m$ such that for any positive integer $n$, $b_{n} \geqslant b_{m}$? If it exists, find the value of $m$; if not, explain the reason.
2. When $n \geqslant 2$,
$$
S_{n}=\frac{a}{1-a}\left(1-a_{n}\right), S_{n-1}=\frac{a}{1-a}\left(1-a_{n-1}\right) .
$$

Then $a_{n}=S_{n}-S_{n-1}$
$$
\begin{array}{l}
=\frac{a}{1-a}\left[\left(1-a_{n}\right)-\left(1-a_{n-1}\right)\right] \\
=\frac{a}{1-a}\left(a_{n-1}-a_{n}\right),
\end{array}
$$

i.e., $a_{n}=a a_{n-1}$.
Since $a_{1}=a \neq 0$, thus, $\left\{a_{n}\right\}$ is a geometric sequence with the first term and common ratio both being $a$.
Therefore, $a_{n}=a^{n}, b_{n}=a_{n} \lg \left|a_{n}\right|=n a^{n} \lg |a|$.
Also, $a=-\frac{\sqrt{7}}{3} \in(-1,0)$, then $\lg |a|<0$.

It can be seen that if there exists a positive integer $m$ that satisfies the condition, then $m$ must be even.
Notice that
$$
\begin{array}{l}
b_{2 k+2}-b_{2 k}=\left[(2 k+2) a^{2 k+2}-2 k a^{2 k}\right] \lg |a| \\
=2 a^{2 k}\left[(k+1) a^{2}-k\right] \lg |a| \\
=2 a^{2 k}\left[(k+1) \times \frac{7}{9}-k\right] \lg |a| \\
=\frac{2 a^{2 k}}{9}(7-2 k) \lg |a|\left(k \in \mathbf{N}_{+}\right) . \\
\text {Also, } \frac{a^{2}}{1-a^{2}}=\frac{7}{2}, \text { thus, }
\end{array}
$$

When $2 k>7$, $b_{2 k+2}>b_{2 k}$, i.e.,
$$
b_{8}<b_{10}<b_{12}<\cdots \text {; }
$$

When $2 k<7$, $b_{2 k+2}<b_{2 k}$, i.e.,
$$
b_{8}<b_{6}<b_{4}<b_{2} \text {. }
$$

Therefore, there exists a positive integer $m=8$, such that for any positive integer $n$, $b_{n} \geqslant b_{m}$.