1. $b_{n+1}=\frac{1}{a_{n+1}-1}=\frac{1}{1-\frac{1}{a_n}}=\frac{a_n}{a_n-1}$. Since $b_n=\frac{1}{a_n-1}$, we have $b_{n+1}-b_n=\frac{a_n}{a_n-1}-\frac{1}{a_n-1}=1$ for all $n\in\mathbb{N}^*$. Therefore, ${b_n}$ is an arithmetic sequence with first term $b_1=\frac{1}{a_1-1}=1$ and common difference $1$.

2. From part 1, we know that $b_n=n$. Hence, $\frac{1}{3}b_n=\frac{1}{3}n$, and $S_n=\frac{1}{3}(1+2+...+n)=\frac{n(n+1)}{6}$. Thus, $\frac{1}{S_n}=\frac{6}{n(n+1)}=6(\frac{1}{n}-\frac{1}{n+1})$. Consequently, $\frac{1}{S_1}+\frac{1}{S_2}+...+\frac{1}{S_n}=6(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1})=6(1-\frac{1}{n+1})=\boxed{\frac{6n}{n+1}}$.

3. We will prove by induction that $T_n<\frac{3}{4}$ for all $n\in\mathbb{N}^*$. From part 1, we have $(\frac{1}{3})^n\cdot b_n=n\cdot(\frac{1}{3})^n$. Hence, $T_n=1\cdot\frac{1}{3}+2\cdot(\frac{1}{3})^2+...+n\cdot(\frac{1}{3})^n$. Multiplying both sides by $\frac{1}{3}$, we get $\frac{1}{3}T_n=1\cdot(\frac{1}{3})^2+2\cdot(\frac{1}{3})^3+...+(n-1)\cdot(\frac{1}{3})^n+n\cdot(\frac{1}{3})^{n+1}$. Subtracting this from the previous equation, we obtain $\frac{2}{3}T_n=\frac{1}{3}+(\frac{1}{3})^2+(\frac{1}{3})^3+...+(\frac{1}{3})^n-n\cdot(\frac{1}{3})^{n+1}=\frac{1}{2}[1-(\frac{1}{3})^n]-n\cdot(\frac{1}{3})^{n+1}$. Therefore, $T_n=\frac{3}{4}-\frac{1}{4}(\frac{1}{3})^{n-1}-\frac{n}{2}\cdot(\frac{1}{3})^n<\boxed{\frac{3}{4}}$.