I will prove $p(n):\ $Any $n$-cent postage where $n \ge 12$ can be made up using $3$-cents and $7$-cents stamps.
My proof:(simple induction)
Base case: as $12= 3+3+3+3$. So it can be made using $3$-cent.
Inductive case: I am assuming that $n$ postage can be made using $3$-cent and $7$-cent, so I will proof that $(n+1)$ can be made using $3$-cent and $7$-cent.
As $n$ postage can be made using $3$-cent and $7$-cent, we can construct $(n+7)$-postage. Then we can construct$((n+7)-3)$-postage, then $((n+7)-3)-3)=(n+1)$.
For instance, $20$-postage can be made using $(7+7+3+3)$, so $((20+1)=(20+7)-3)-3)$.
I know it is may be wrong. But I can not realize why?
Another question is, how many base case is needed for strong induction? I don't know. Please explain be done by anyone.
The problem with your proof is you are removing 3 cent stamps. Your initial case only contains four lots of 3 cent stamps. So after your inductive step occurs two times you have no 3 cent stamps left and you can not continue indefinitely. More generally you can not guarantee that any arbitrary case $n$ will have a 3 cent stamp to remove.
A simpler approach would be to have 3 initial cases for 12,13 and 14 and have an inductive step of add a 3 cent stamp.
Edit: If you don't want to have three bases and do it the easy way you need to demonstrate that every $n$ case is one of two cases and have two different inductive steps:
The $n$ case has (at least) two 3 cent stamps and you can then increase by one like you describe.
The $n$ case has (at least) two 7 cent stamps and you can then increase by one by removing two 7 cent stamps and adding five 3 cent stamps.
To show that you always have one of these two cases is significantly harder.