1. Let 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 set $A=$ $\qquad$
According to the problem, without loss of generality, let $a_{1}<a_{2}<a_{3}<a_{4}$, then we have
$$
\left\{\begin{array}{l}
a_{1}+a_{2}+a_{3}=-1, \\
a_{1}+a_{2}+a_{4}=3, \\
a_{1}+a_{3}+a_{4}=5, \\
a_{2}+a_{3}+a_{4}=8
\end{array} \Rightarrow a_{1}+a_{2}+a_{3}+a_{4}=5 \Rightarrow\left\{\begin{array}{l}
a_{1}=-3, \\
a_{2}=0, \\
a_{3}=2, \\
a_{4}=6
\end{array} \Rightarrow A=\{-3,0,2,6\} .\right.\right.
$$