1. For $n\geqslant 2$, from the system of equations $\begin{cases}S_n-2S_{n-1}=1 \\ S_{n+1}-2S_n=1\end{cases}$, by subtracting the two equations, we get $a_{n+1}-2a_n=0$.
When $n=2$, $a_2=2$,
thus $\frac{a_{n+1}}{a_n}=2$ for $n\in\mathbb{N}^*$,
which means $\{a_n\}$ is a geometric sequence with the first term $1$ and common ratio $2$.
2. From (1), we have $a_n=2^{n-1}$, thus $c_n=n\times\left(\frac{1}{2}\right)^{n-1}$,
therefore, $T_n=1\times\left(\frac{1}{2}\right)^0+2\times\left(\frac{1}{2}\right)^1+3\times\left(\frac{1}{2}\right)^2+\ldots+(n-1)\times\left(\frac{1}{2}\right)^{n-2}+n\times\left(\frac{1}{2}\right)^{n-1}$
Hence, $\frac{1}{2}T_n=1\times\left(\frac{1}{2}\right)^1+2\times\left(\frac{1}{2}\right)^2+3\times\left(\frac{1}{2}\right)^3+\ldots+(n-1)\times\left(\frac{1}{2}\right)^{n-1}+n\times\left(\frac{1}{2}\right)^n$
Subtracting the two expressions, we get $\frac{1}{2}T_n=\left(\frac{1}{2}\right)^0+\left(\frac{1}{2}\right)^1+\left(\frac{1}{2}\right)^2+\ldots+\left(\frac{1}{2}\right)^{n-1}-n\times\left(\frac{1}{2}\right)^n=2-(n+2)\times\left(\frac{1}{2}\right)^n$
Therefore, $T_n=4-(n+2)\times\left(\frac{1}{2}\right)^{n-1} 0$, we have $T_n\geqslant T_1=1$,
In conclusion, there exist positive integers $M$ and $m$ such that $m\leqslant T_n < M$ holds for any positive integer $n$, where $m=1$ and $M\geqslant 4$ with $M\in\mathbb{N}$.
Thus, the final answers are $\boxed{m=1}$ and $\boxed{M\geqslant 4}$ with $M\in\mathbb{N}$.