1. Define an operation “ * ”: For any pair of real numbers $(x, y)$, it always holds that
$$
(x, y) *(x, y)=\left(x+y+1, x^{2}-y-1\right) \text {. }
$$

If real numbers $a$ and $b$ satisfy $(a, b) *(a, b)=(b, a)$, then $a=$ $\qquad$ ,$b=$ $\qquad$ .
$$
\text { II. 1. }-1,1 \text {. }
$$

According to the problem, we get
$$
\left\{\begin{array}{l}
a+b+1=b, \\
a^{2}-b-1=a .
\end{array} \text { Solving, we get } a=-1, b=1\right. \text {. }
$$