1) 1, 1
2) -1, -1
3) 1, – 1
4) -1, 1
Solution: (4) -1, 1
\(\begin{array}{l}\begin{array}{l} \text { Given } \int_{a}^{b} x^{3} d x=0 \\ \Rightarrow\left[\frac{x^{4}}{4}\right]_{a}^{b}=0 \\ \Rightarrow b^{4}-a^{4}=0 \\ \Rightarrow b^{2}-a^{2}=0 \\ \Rightarrow(b+a)=0 \text { or } b-a=0 \\ \Rightarrow b=-a\{\text { Since } b \neq \text { a given }\} \\ \therefore \int_{a}^{b} x^{2} d x=\frac{2}{3} \\ \Rightarrow \int_{a}^{-a} x^{2} d x=\frac{2}{3} \\ \Rightarrow\left[\frac{x^{3}}{3}\right]_{a}^{-a}=\frac{2}{3} \\ \Rightarrow \frac{-a^{3}-a^{3}}{3}=\frac{2}{3} \\ \Rightarrow \frac{-2 a^{3}}{3}=\frac{2}{3} \\ \Rightarrow a^{3}=1 \\ a=-1 \text { and } b=1 \end{array}\end{array} \)