解：??$(1)$??由于??$30=3×1+27$??
設(shè)??$2g$??的砝碼有??$x$??個，則??$5g$??的砝碼有??$(15-3-x)$??個
??$27=2x+5(15-3-x)$??
解得，??$x=11$??
??$15-3-11=1$??
所以??$2g$??的砝碼有??$11$??個，??$5g$??的砝碼有??$1$??個
??$(2)$??設(shè)??$1g$??的砝碼有??$a$??個，??$2g$??的砝碼有??$b$??個，則??$5g$??的砝碼有??$(15-a-b)$??個
根據(jù)題意，得??$30=a+2b+5(15-a-b)=a+2b+75-5a-5b$??
則??$4a+3b=45$??
所以??${{\begin{cases} { {a=9，}} \\{b=3，}\end{cases}}}{{\begin{cases} { {a=6，}} \\{b=7，}\end{cases}}}{{\begin{cases} { {a=3，}} \\{b=11，}\end{cases}}}$??
故除第??$(1)$??小題的情況外，取出的砝碼數(shù)量如下：
??$①1g$??的砝碼有??$9$??個，??$2g$??的砝碼有??$3$??個，??$5g$??的砝碼有??$3$??個;
??$②1g$??的砝碼有??$6$??個，??$2g$??的砝碼有??$7$??個，??$5g$??的砝碼有??$2$??個.
