related to 2nd part is 2018 Shortlist N1 , solved here 2018 Shortlist N1 wrote: Determine all pairs $(n, k)$ of distinct positive integers such that there exists a positive integer $s$ for which the number of divisors of $sn$ and of $sk$ are equal.

For the first part, just notice that product of all divisors of a number $t$ is $t^{ \frac{\tau(t)}{2}}$, for which $\tau(t)$ denote the number of divisors of $t$ Then if product of all divisor of $2$ numbers $a,b$ are equal means $$a^{ \frac{\tau(a)}{2}}=b^{ \frac{\tau(b)}{2}}$$Which means $a$ and $b$ shares the same set of prime divisors Now just take $p \mid a,b$ and concentrate on the exponent $$ \frac{\tau(a)}{2} v_p(a) = \frac{\tau(b)}{2} v_p(b)$$Suppose that $a \ne b$, which means there exists $p$ such that $v_p(a) \ne v_b(b)$, WLOG suppose that $v_p(a) < v_b(b)$ Then $\frac{{\tau (a)}}{2} > \frac{{\tau (b)}}{2} \Leftrightarrow \tau (a) > \tau (b) \implies {v_p}(a) < {v_p}(b) $ $\forall p|a,b$ Then \[\tau (a) = \prod {({v_p}(a) + 1)} < \prod {({v_p}(b) + 1)} = \tau (b)\]which is a contradiction. Then if product of all divisor of $2$ numbers $a,b$ are equal means $a=b$ Then if there exists $s$ then $m=n$