Finding divisibility of a

Since $71$ is prime, $72!$ is divisible by $71$ whereas $36!^2$ is not. Hence $72!/(36!^2)$ is divisible by $71$ and $72!/(36!^2)-1$ is not.

For divisibility by $73$, notice that $$36!^2 = \prod_{k=1}^{36} k^2 \equiv \prod_{k=1}^{36} (-k(73-k)) = \prod_{k=1}^{36} k(73-k) = 72! \mod 73,$$ so the quotient $72!/(36!^2)$ is $1$ modulo $73$ (note that $73$ is prime, thus $72!$ is not divisible by $73$). Since $72!/(36!^2) \equiv 1 \mod 73$, we find that $72!/(36!^2)-1$ is divisible by $73$.