1. Set $A=\left\{1, \frac{x+y}{2}-1\right\}$, Set $B=\{-\ln (x y), x\}$, if $A=B$ and $0<y<2$, then $\left(x^{2}-\frac{1}{y^{2}}\right)+\left(x^{4}-\right.$ $\left.\frac{1}{y^{4}}\right)+\cdots+\left(x^{2022}-\frac{1}{y^{2022}}\right)$ is $\qquad$
Answer: 0
Solution: Given $A=B$, if $1=x$, substituting into $\frac{x+y}{2}-1=-\ln (x y)$, it is easy to get $y=1$, at this point the result of the required expression is 0; if $\frac{x+y}{2}-1=x$, we get $y(y-2)=\frac{1}{e}$, but $0<y<2, y(y-2)<0$, which is a contradiction. In summary, the required value is 0.