1. Find natural numbers $a, b, c$, such that for any $n \in \mathbf{N}, n>2$, we have
$$
b-\frac{c}{(n-2)!}<\frac{2^{3}-a}{2!}+\frac{3^{3}-a}{3!}+\cdots+\frac{n^{3}-a}{n!}<b .
$$
1. Let $b_{n}=b-\frac{2^{3}-a}{2!}-\frac{3^{3}-a}{3!}-\cdots-\frac{n^{3}-a}{n!}=\frac{P(n)}{n!}$, where $P(n)$ is a polynomial in $n$. Clearly, the degree of $P(n)$ cannot be greater than 2. Set $P(n)=k n^{2}+l n+m$.
By $b_{n+1}-b_{n}=-\frac{(n+1)^{3}-a}{(n+1)!}=\frac{P(n+1)}{(n+1)!}-\frac{P(n)}{n!}$
and the method of undetermined coefficients, we get $k=1, l=3, m=5, a=5$.
Thus, $b=\frac{2^{3}-5}{2!}+\frac{3^{3}-5}{3!}+\cdots+\frac{n^{3}-5}{n!}+\frac{n^{3}+3 n+5}{n!}(n>2)$.
Taking $n+1$ and $n$ in the above equation and subtracting the two, we find that $b$ is a constant. Setting $n=3$, we get $b=9$.
By $b_{n}(n-2)!\cdot b_{n}=(n-2)!\cdot \frac{n^{2}+3 n+5}{n!}=1+\frac{4 n+5}{n^{2}-n}$.

Noting that $\frac{4 n+5}{n^{2}-n}$ is a decreasing function of $n$ for $n>2$, we have $c>1+\frac{4 \times 3+5}{3^{2}-3}=\frac{23}{6}$.
We can take $c=4$. Thus, $a=5, b=9, c_{\text {min }}=4$.
Note (1) Using a similar method, we can consider more general cases: Determine natural numbers $a, b, c$ such that for any $n, k \in \mathbf{N}, n>k-1$, we have $b-\frac{c}{(n-k+1)!}2)$.
$2^{\circ} \quad \frac{2^{3}-5}{2!}+\frac{3^{3}-5}{3!}+\cdots+\frac{n^{3}-5}{n!}+\frac{n^{2}+3 n+5}{n!}=9(n>2)$.
$3^{\circ} \quad \frac{2^{5}-52}{2!}+\frac{3^{5}-52}{3!}+\cdots+\frac{n^{5}-52}{n!}+\frac{n^{4}+5 n^{3}+14 n^{2}+31 n+52}{n!}=103$.

These results provide rich material for mathematics competitions.