How to draw NPDA for words whose number of b's is strictly more than that of a's but strictly less than twice the amount I know that CFG for $$ \{a^{m}b^{n}\mid m\leq n\leq 2m \}$$ is
$$ S\rightarrow ab/abb/aSb/aSbb $$ but I am not able to tweak  it in such a way that it is strictly in between m and 2m and not equal to any of m or 2m. 
I want to draw PDA for the same.

For drawing the PDA what I did was to push non-deterministically  1 a and 2a's and popping them out for b one by one. How should I handle the edge case that it always doesn't push 1 a or not always 2 a's?
As you knew intuitively, you just missed the solution by a millimeter, which, however, may look like a mile away.

$S\rightarrow aabbb/aSb/aSbb$

The idea is simple. Since $ab$ and $abb$ should not be generated, what is the shortest or simplest word that can be generated?

Here is a sketch of a NPDA. When it reads the first $a$, it goes to state $q_1$. When it reads the second $a$, it goes to state $q_2$. Then every time if it reads one $a$, it will push one or two $a$'s nondeterministically onto the stack. Then when it reads the first $b$ it goes to state $q_4$.  When it reads the second $b$ it goes to state $q_5$. When it reads the third $b$ it goes to state $q_6$. Then every time if it reads one $b$, it pops out one $a$ from the stack, staying at state $q_6$. 

I will let you flesh out the details.