For a) Just take $6$ and $11$, so Yes. For b) Answer is No. $\textsf{Proof: }$ Consider $2$ natural numbers $a$ and $b$. Product of divisors of $a= \sqrt{a^{d(a)}}=a^{\frac{d(a)}{2}}$, where $d(n)$ is the no.of divisors of $n$. Product of divisors of $b= \sqrt{b^{d(b)}}=b^{\frac{d(b)}{2}}$. As $a^{\frac{d(a)}{2}}=b^{\frac{d(b)}{2}}$, we can say the the set of their prime divisors is same. Consider an arbitrary prime $p$ which divides both $a$ and $b$. Again, as $a^{\frac{d(a)}{2}}=b^{\frac{d(b)}{2}}$, the exponent of $p$ on both sides is equal. $\implies \nu_p(a) \cdot \frac{d(a)}{2}=\nu_p(b) \cdot \frac{d(b)}{2} \implies \frac{\nu_p(a)}{ \nu_p(b)}= \frac{d(a)}{d(b)}$ So, as this is true for all such primes,it implies that $a=b$. (since if $d(a) >d(b)$ or vice-versa, the obtained relation is rendered false).