1. Given that $P(n, m) = \sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\frac{m}{m+k}$ and $Q(n, m) = \binom{n+m}{m}$, and $m,n\in\mathbb{N}^*$; when $m=1$,

$P(n, 1) = \sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\frac{1}{1+k}$ can be rewritten using the identity $\binom{n+1}{k+1} = \binom{n}{k} + \binom{n}{k+1}$, which simplifies to $\frac{1}{n+1}$.

$Q(n, 1) = \binom{n+1}{1} = n+1$.

Therefore, $P(n,1)\cdot Q(n,1) = \boxed{1}$.

2. Let's prove that $P(n, m)\cdot Q(n, m)$ is a constant for all $m \in \mathbb{N}^*$.

Starting with $P(n, m) = \sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\frac{m}{m+k}$, which can be decomposed into two parts:

$P(n, m) = 1 - \sum_{k=1}^{n-1}(-1)^{k}\left(\binom{n-1}{k} + \binom{n-1}{k-1}\right)\frac{m}{m+k} + (-1)^{n}\frac{m}{m+n}$.

The first summation telescopes and using the identity $\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}$, we get:

$P(n, m) = P(n-1, m) + \frac{m}{n}\left[\sum_{k=1}^{n}(-1)^{k}\binom{n}{k}\left(1 - \frac{m}{m+k}\right)\right]$.

Simplifying further by distributing the summation and collecting terms:

$P(n, m) = P(n-1, m) + \frac{m}{n}\left[-1 - \sum_{k=1}^{n}(-1)^{k}\binom{n}{k}\frac{m}{m+k}\right]$.

Hence, we obtain a recursive relationship:

$P(n, m) = P(n-1, m) - \frac{m}{n} P(n, m)$.

From the recursion, we can infer that:

$P(n, m) = \frac{n! \cdot m!}{(n+m)!} P(0, m) = \frac{1}{\binom{n+m}{n}}$.

Since $Q(n, m) = \binom{n+m}{n}$, we find that:

$P(n, m) \cdot Q(n, m) = \boxed{1}$, demonstrating that it is indeed a constant value.