$(2)解:由(1)知这列数以-\frac{1}{3},\frac{3}{4},4为一个循环组，依 次出现\ $
$因为-\frac{1}{3}+\frac{3}{4}+4=-\frac{4}{12}+\frac{9}{12}+\frac{48}{12}=\frac{53}{12},2022÷3=674$
$所以a_1+a_2+a_3+...+a_{2022}=(a_1+a_2+a_3)+…+(a_{2020}+a_{2021}+a_{2022})\ $
$=\frac{53}{12}×674\ $
$=\frac{17861}{6}$
