I'll assume here that the number $2M5$ and $13N$ are concatenated three-digit numbers rather than products, because if they are products then the question is trivial.
Being divisible by 36 is equivalent to
being divisible by both of its prime factors $4=2^2$ and $9=3^2$.
$2M5$ is odd, so both factors of 2 must come from $13N$. That means $3N$ must be a multiple of 4, so either $N=2$ or $N=6$. In either of these cases, the product is definitely a multiple of 4 - it's an "if and only if" condition.
If the product is divisible by 9, then either both factors are multiples of 3 (i.e. their digit sums are multiples of 3) or one of the two factors is a multiple of 9 (i.e. its digit sum is a multiple of 9).
Let's consider the two cases from the first numbered point above:
If $N=2$, then $132$ is divisible by 3 but not 9, so we need $2M5$ divisible by 3, i.e. $2+M+5$ divisible by 3, which means either $M=2$ or $M=5$ or $M=8$. In all of these cases, the product is definitely a multiple of 9.
If $N=6$, then $136$ is not divisible by 3, so we need $2M5$ divisible by 9, i.e. $2+M+5$ divisible by 9, which means $M=2$. Again, in this case the product is definitely a multiple of 9.
So the possibilities for the pair $(M,N)$ are:
$(2,2),(5,2),(8,2),(2,6)$ - four possibilities in all.