解:$(1)\{\begin{array}{l}2023x+2022y=2021①\\2021x+2020y=2019②\end{array}$
①-②得$ 2x+2y=2,$ 即$ x+y=1③$
$③×2020$得$ 2020x+2020y=2020④$
②-④得$ x=-1$
把$ x=-1$代入③得$ -1+y=1,$解得$ y=2$
∴原方程组的解是$ \{\begin{array}{l}x=-1\\y=2\end{array}$
$(2) \{\begin{array}{l}(a+2)x+(a+1)y=a ①\\(b+2)x+(b+1)y=b②\end{array}$
①-②得$(a-b)x+(a-b)y=a-b$
∵$a \neq b$
∴$a-b \neq 0$
∴$x+y=1③$
$③×(a+1)$得$ (a+1)x+(a+1)y=a+1④$
①-④得$ x=-1$
把$ x=-1 $代入③得$ -1+y=1,$ 解得$ y=2$
∴原方程组的解是$ \{\begin{array}{l}x=-1\\y=2\end{array}$