a) Yeah just look at the Wikipedia table of values for $\sigma _1 (x)$ (the first-degree divisor function) and find some easy example like $14$ and $15$, both of which have a divisor sum of $24$. b) It's well known that the product of positive divisors of $n$ is given by $n^{\sigma _0 (n)/2}$. Thus, we want to show that for $n,m \in \mathbb{Z},$ $$n^{\sigma _0 (n)/2} = m^{\sigma _0 (m)/2} \implies n = m.$$Consider an arbitrary prime $p|n$ and let $a = \nu _p (n)$ and $b = \nu _p (m)$ be $p$-adic evaluations. Given the above, it follows that $$p ^{a \cdot \sigma _0 (n)/2} = p^{b \cdot \sigma _0 (m) /2} \implies \frac{a}{b} = \frac{\sigma _0 (m)}{\sigma _0 (n)}.$$Do case-work on $\sigma _0(n) < \sigma_0 (m)$ and $\sigma _0(n) > \sigma_0 (m)$ for some obvious contradictions and the conclusion $\sigma _0(n) = \sigma_0 (m)$ follows. Since $p$ was chosen arbitrarily, we conclude that $n = m$.

For (b), you could also pair off the factors of a number $n$, making a special case if $n$ is square.