3)
(a)
m€Z,
7-m≥0 => m≤7
2m-8≥0 => m≥4
m€{4;5;6;7}
(b)
m€Z
m>0
3m-9≥0
(6/m)€Z
m€{3;6}
(c)
3-(8/m)≥0=> (3m-8)≥0
(because m>0) m≥8/3=2⅔
m+2≥0
8/m€Z
m>0
m€{4,8}
4)
(3a-6)x³+(5-b)x+15
if it's a constant polynomial
3a-6=0
a=2
5-b=0
b=5
a+b=2+5=7
Good luck!