(a)  the last, ones digit, is any of 5 options 0, 2, 4, 6, 8.

     the tens digit is  any from 0 to 9 : 10 options;

     the leading digit: any from 1 to 9 :  9 options.


     Total amount of such numbers is  9*10*5 = 450.


     The same answer you can obtain by different way
         noticing that there are 900 three-digit numbers from 100 to 999,
         and exactly half of them are even numbers.




(b)  If a three-digit number is divisible by 5, its ones digit must be 0 or 5.

     So, the amount of such three-difit numbers is 9*10*2 = 180.


     Again, it is exactly    part of the total 900 three-digit numbers.



(c)  Let me be short in this last case:


     The amount of the three-digit numbers divisible by 10 is exactly    part 
     of the total 900 three-digit numbers, i.e. 90, in total.