$N=2^6*3^4*5^6=A*B$
In how many ways N can be written as a product of two numbers that are even?
$N=2A*2B$
then how $2^4*3^4*5^6=a*b$? and why not $2^6*3^4*5^6=a*b$
With the expression $2^4*3^4*5^6=a*b$ the number of factors was computed as $5\times 5\times 7=175$. Then the number of pairs was calculated as $\frac {175-1}2+1=88$. How does this calculation make sense?
In answer to your question $A,B$ and $a,b$ are both ambiguously defined and are likely confusing you.
If $$N=2^6\cdot 3^4\cdot 5^6=A\cdot B$$ is the product of even factors, we can write$A=2a, B=2b$ so that $$N=2a\cdot 2b$$ and we have $$a\cdot b=2^4\cdot 3^4\cdot 5^6=\frac A2\cdot \frac B2=\frac N4$$
There are then $175$ factors. One of these is the square root and the others come in pairs $p\cdot q=\frac N4$. So if we want the number of pairs, we take off one for the square root, and have $\frac {174}2=87$ pairs (here we are counting the factorisation $p\cdot q$ as being the same as $q\cdot p$). To get all the factorisations we add back the one which involves the square root - this being $p^2$ involves just one of the $175$ factors - and obtain the answer $88$.