1. For a finite set $A$, there exists a function $f: \mathbf{N}^{*} \rightarrow A$, with the following property: if $i, j \in \mathbf{N}^{*}$, and $|i-j|$ is a prime number, then $f(i) \neq f(j)$. How many elements must the set $A$ have at least?
1. Since the absolute value of the difference between any two of the numbers $1,3,6,8$ is a prime number, according to the problem, $f(1)$, $f(3)$, $f(6)$, $f(8)$ are four distinct elements in $A$. Therefore, $|A| \geqslant 4$. On the other hand, if we let $A=\{0,1,2,3\}$, and the mapping $f: \mathbf{N}^{*} \rightarrow A$ is defined as: if $x \in \mathbf{N}^{*}, x=4 k + r$, then $f(x)=r$, where $k \in \mathbf{N}, r=0,1,2,3$, then for any $x, y \in \mathbf{N}^{*}$, if $|x-y|$ is a prime number, assuming $f(x) = f(y)$, then $x \equiv y(\bmod 4)$, thus $4 \mid |x-y|$, which contradicts the fact that $|x-y|$ is a prime number. Hence, the set $A$ contains at least 4 elements.