1. Let the set $A=\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}$. If the set $B=\{-1,3,5,8\}$ consists of the sums of all three-element subsets of $A$, then the set $A=$ $\qquad$ .
- 1. $\{-3,0,2,6\}$.

Obviously, in all three-element subsets of set $A$, each element appears 3 times. Therefore,
$$
\begin{array}{l}
3\left(a_{1}+a_{2}+a_{3}+a_{4}\right)=(-1)+3+5+8=15 \\
\Rightarrow a_{1}+a_{2}+a_{3}+a_{4}=5 .
\end{array}
$$

Thus, the four elements of set $A$ are
$$
5-(-1)=6,5-3=2,5-5=0,5-8=-3 \text {. }
$$

Therefore, set $A=\{-3,0,2,6\}$.