To prove:
(1) Since $a_{n+2}=3a_{n+1}-2a_n$ for $n\in\mathbb{N}^*$,
we have $a_{n+2}-a_{n+1}=2(a_{n+1}-a_n)$ for $n\in\mathbb{N}^*$.
Because $a_2-a_1=3-1=2$,
the sequence $\{a_{n+1}-a_n\}$ is a geometric sequence with both initial term and common ratio equal to 2.

(2) From (1) we know that $a_{n+1}-a_n=2^n$, and clearly, the sequence $\{a_n\}$ is increasing, therefore
$$b_n=\frac{2^{n-1}}{a_n\cdot a_{n+1}}=\frac{1}{2}\cdot\frac{2^n}{a_n\cdot a_{n+1}}=\frac{1}{2}\cdot\frac{a_{n+1}-a_n}{a_n\cdot a_{n+1}}=\frac{1}{2}\left(\frac{1}{a_n}-\frac{1}{a_{n+1}}\right),$$
which leads to
$$T_n=\frac{1}{2}\left(\frac{1}{a_1}-\frac{1}{a_2}+\frac{1}{a_2}-\frac{1}{a_3}+\dots+\frac{1}{a_n}-\frac{1}{a_{n+1}}\right)$$
$$=\frac{1}{2}\left(\frac{1}{a_1}-\frac{1}{a_{n+1}}\right)$$
$$=\frac{1}{2}\left(1-\frac{1}{a_{n+1}}\right)$$
$$<\frac{1}{2}.$$
Hence, we conclude that $\boxed{T_n<\frac{1}{2}}$ for all $n$.