解：$(2)$由$(1)$知这列数以$-\frac {1}{3}$，$\frac {3}{4}$，$4$为一个循环，依次出现
∵$-\frac {1}{3}+\frac {3}{4}+4=-\frac {4}{12}+\frac {9}{12}+\frac {48}{12}=\frac {53}{12}$
$2022÷3=674$
∴$a_{1}+a_{2}+a_{3}+···+a_{2022} =(a_{1}+a_{2}+a_{3})+···+(a_{2020}+a_{2021}+a_{2022}) $
$=\frac {53}{12}×674 =\frac {17861}{6}$
