解：?$(1)$?正確，原因是一個(gè)數(shù)的倒數(shù)的倒數(shù)等于其自身；
?$(2)$?原式的倒數(shù)為?$(\frac {1}{3}-\frac {1}{6}+\frac {3}{8})÷(-\frac {1}{24})$?
?$=(\frac 13-\frac 16+\frac 38)×(-24)$?
?$=-24×\frac {1}{3}+24×\frac {1}{6}-24×\frac {3}{8}$?
?$=-8+4-9$?
?$=-13$?
則?$(-\frac {1}{24})÷(\frac {1}{3}-\frac {1}{6}+\frac {3}{8})=-\frac {1}{13}$?
