1. Proof: Given $S_{n+1}=4a_n+2$ and $S_{n+2}=4a_{n+1}+2$, subtracting these two equations yields $S_{n+2}-S_{n+1}=4(a_{n+1}-a_n)=4a_{n+1}-4a_n$. Therefore, $a_{n+2}-2a_{n+1}=2(a_{n+1}-2a_n)$. Since $b_n=a_{n+1}-2a_n$, we have $b_{n+1}=2b_n$. Given $1+a_2=4+2=6$, we find $a_2=5$. Thus, $b_1=a_2-2a_1=3$. Therefore, the sequence $\{b_n\}$ is a geometric sequence with the first term $3$ and common ratio $2$, so $\boxed{\text{The sequence } \{b_n\} \text{ is a geometric sequence with the first term } 3 \text{ and common ratio } 2}$.

2. Proof: From (1), we have $b_n=3\cdot 2^{n-1}$. Therefore, $b_n=a_{n+1}-2a_n=3\cdot 2^{n-1}$. This leads to $\frac {a_{n+1}}{2^{n+1}}- \frac {a_n}{2^n}= \frac {3}{4}$, which means $c_{n+1}-c_n= \frac {3}{4}$. Therefore, the sequence $\{c_n\}$ is an arithmetic sequence with the first term $\frac {1}{2}$ and common difference $\frac {3}{4}$, so $\boxed{\text{The sequence } \{c_n\} \text{ is an arithmetic sequence with the first term } \frac {1}{2} \text{ and common difference } \frac {3}{4}}$.

3. Solution: From (2), we have $c_n= \frac {1}{2}+ \frac {3}{4}(n-1)= \frac {3}{4}n- \frac {1}{4}$. This implies $\frac {a_n}{2^n}= \frac {3}{4}n- \frac {1}{4}$, therefore $a_n=(3n-1)2^{n-2}$. The sum of the first $n$ terms $S_n=4a_{n-1}+2=(3n-4)\cdot 2^{n-1}+2$. Thus, the general term formula for the sequence $\{a_n\}$ is $\boxed{a_n=(3n-1)2^{n-2}}$ and the sum of its first $n$ terms is $\boxed{S_n=(3n-4)\cdot 2^{n-1}+2}$.