Answer
a) 9
b) 6
c) 2
Work Step by Step
Note that neither 5 nor 31 is divisible by either 3 or 4.
a) There are $\left \lfloor{\frac{31}{3}}\right \rfloor$=10 integers less than 31 that are divisible by 3, and $\left \lfloor{\frac{5}{3}}\right \rfloor$=1 of them are less than 5 as well. This leaves 10-1=9 numbers between 5 and 31 that are divisible by 3. They are 6, 9, 12, 15, 18, 21, 24, 27 and 30.
b) There are $\left \lfloor{\frac{31}{4}}\right \rfloor$=7 integers less than 31 that are divisible by 4, and $\left \lfloor{\frac{5}{4}}\right \rfloor$=1 of them are less than 5 as well. This leaves 7-1=6 numbers between 5 and 31 that are divisible by 3. They are 8, 12, 16, 20, 24 and 28.
c) A number is divisible by both 3 and 4 if and only if it is divisible by their least common multiple, which is 12. Obviously there are only two such number between 5 and 31, namely 12 and 24.
This could also be done like previous parts : $\left \lfloor{\frac{31}{12}}\right \rfloor$-$\left \lfloor{\frac{5}{12}}\right \rfloor$=2-0=2.
This can also be seen as intersection of integers obtained in part a and b.