12. (15 points) Given a sequence of positive numbers $\left\{a_{n}\right\}$, the sum of the first $n$ terms is $S_{n}$, and it satisfies $a_{n}^{2}+a_{n}-2 S_{n}=0$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) If $b_{1}=1,2 b_{n}-b_{n-1}=0\left(n \geqslant 2, n \in \mathbf{Z}_{+}\right)$, let $c_{n}=a_{n} b_{n}$, find the sum of the first $n$ terms of the sequence $\left\{c_{n}\right\}$, denoted as $T_{n}$;
(3) Do there exist integers $m, M$ such that $m<T_{n}<M$ holds for any positive integer $n$, and $M-m=4$? Please explain your reasoning.
Three, 12. (1) From the given,
$$
\begin{array}{l}
a_{n-1}^{2}+a_{n-1}-2 S_{n-1}=0 \\
\Rightarrow\left(a_{n}-a_{n-1}\right)\left(a_{n}+a_{n-1}\right)+a_{n}-a_{n-1}-2 a_{n}=0 \\
\Rightarrow\left(a_{n}+a_{n-1}\right)\left(a_{n}-a_{n-1}-1\right)=0 \\
\Rightarrow a_{n}-a_{n-1}=1 .
\end{array}
$$

Let $n=1$.
Then $a_{1}^{2}+a_{1}-2 a_{1}=0 \Rightarrow a_{1}=1$.
Thus, $a_{n}=1+(n-1)=n$.
(2) Note that, $\frac{b_{n}}{b_{n-1}}=\frac{1}{2}$.

Therefore, $b_{n}=\left(\frac{1}{2}\right)^{n-1}$.
Hence, $c_{n}=n\left(\frac{1}{2}\right)^{n-1}$.
Then $T_{n}=\left(\frac{1}{2}\right)^{0}+2\left(\frac{1}{2}\right)^{1}+\cdots+n\left(\frac{1}{2}\right)^{n-1}$
$$
\Rightarrow \frac{1}{2} T_{n}=\left(\frac{1}{2}\right)^{1}+2\left(\frac{1}{2}\right)^{2}+\cdots+n\left(\frac{1}{2}\right)^{n} \text {. }
$$

Subtracting the above two equations gives
$$
\frac{1}{2} T_{n}=1+\frac{1}{2}+\cdots+\left(\frac{1}{2}\right)^{n-1}-n\left(\frac{1}{2}\right)^{n} \text {. }
$$

Thus, $T_{n}=4\left[1-\left(\frac{1}{2}\right)^{n}\right]-n\left(\frac{1}{2}\right)^{n-1}$
$$
=4-(2 n+4)\left(\frac{1}{2}\right)^{n} \text {. }
$$
(3) From (2), $T_{n}0,
\end{array}
$$

Then $T_{n} \geqslant T_{1}=1$.
Therefore, there exist integers $M=4, m=0$ that satisfy the problem's requirements.