1) Denote $d(x)$, $\pi(x)$ as the number of divisors, the product of divisors of positive integer $x$.
Firstly, we can see that for any divisor $y_{k}$ of $x$, $1 \leq k \leq d(x)$ then $\frac{x}{y_{i}}$ is also divisor of $x$, thus
$$\prod_{k=1}^{d(x)} y_{k} = \prod_{k=1}^{d(x)} \frac{x}{y_{k}} \text{ so } \left(\prod_{k=1}^{d(x)} y_{k}\right)^{2} = x^{d(x)}, \text{ hence } \pi(x) = x^{\frac{d(x)}{2}}$$
Thus for any $m, n \in \mathbb{Z}^{+}$ then $\pi(m) = \pi(n)$ implies that $m, n$ share the common prime divisors set $S$.
Suppose that $d(m) \geq d(n)$ and take $p \in S$. Since $m^{d(m)} = n^{d(n)}$, we have $d(m) \cdot v_{p}(m) = d(n) \cdot v_{p}(n)$. Since $d(m) \geq d(n)$, we get $v_{p}(m) \leq v_{p}(n)$. And this is true for all $p \in S$, thus
$$d(m) = \prod_{p \in S} (v_{p}(m) + 1) \leq \prod_{p \in S} (v_{p}(n) + 1) = d(n)$$
So $d(m) = d(n)$, which implies that $m = n$. Therefore, we can find a positive integer such that product of all divisors of $s m$, $s n$ are equal if and only if $m = n$.

2) Firstly, we can see that if $n \mid m$ then any divisor of $s n$ is also divisor of $s m$, so $d(s n) < d(s m)$. We consider $n \nmid m$, and denote $p_{1}, p_{2}, \ldots, p_{t}$ be all prime dividing $m n$. Suppose that
$$m = \prod_{i=1}^{t} p_{i}^{\alpha_{i}} \text{ and } n = \prod_{i=1}^{t} p_{i}^{\beta_{i}}$$
Now we are looking for $s = \prod_{i=1}^{t} p_{i}^{\gamma_{i}}$ such that
$$\frac{d(s m)}{d(s n)} = \prod_{i=1}^{t} \frac{\alpha_{i} + \gamma_{i} + 1}{\beta_{i} + \gamma_{i} + 1} = 1$$
Note that if $\alpha_{i} = \beta_{i}$, then regardless of the value of $\gamma_{i}$, the corresponding factor equals to 1 and does not affect the product. So we may assume that $\alpha_{i} \neq \beta_{i}$ for all $1 \leq i \leq t$.

Claim. Let $\alpha > \beta$ be nonnegative integers. Then for every $M \geq \beta + 1$, there exist a nonnegative integer $\gamma$ such that
$$\frac{\alpha + \gamma + 1}{\beta + \gamma + 1} = \frac{M + 1}{M} .$$
It is equivalent to $\gamma = M(\alpha - \beta) - (\beta + 1) \geq 0$, which is true.

Back to the original problem, we can assume that $\alpha_{i} > \beta_{i}$ for $i = 1, 2, \ldots, u$ and $\alpha_{i} < \beta_{i}$ for $i = u + 1, u + 2, \ldots, t$. Take some big enough $X$ and choose $\gamma_{i}$ such that
- $\frac{\alpha_{i} + \gamma_{i} + 1}{\beta_{i} + \gamma_{i} + 1} = \frac{u X + i}{u X + i - 1}$ for $1 \leq i \leq u$.
- $\frac{\beta_{u + i} + \gamma_{u + i} + 1}{\alpha_{u + i} + \gamma_{u + i} + 1} = \frac{(t - u) X + i}{(t - u) X + i - 1}$ for $1 \leq i \leq t - u$.
Then we have
$$\frac{d(s m)}{d(s n)} = \prod_{i=1}^{u} \frac{u X + i}{u X + i - 1} \cdot \prod_{i=1}^{t - u} \frac{(t - u) X + i - 1}{(t - u) X + i} = \frac{u(X + 1)}{u X} \cdot \frac{(t - u) X}{(t - u)(X + 1)} = 1 .$$