There are two cases since 0 is an even digit and it can go last, but it
can't go first.  So the number of choices for the first digit is different
when we choose 0 for the last digit than if we choose 2 or 6 for the last
digit.  So we must break this problem into two cases:

Case 1: 0 goes last

Choose the last (5th) digit as 0 in 1 way.
Choose the 1st digit any of 5 ways. {2,3,5,6,9}
Choose the 2nd digit any of the remaining 4 ways.
Choose the 3rd digit any of the remaining 3 ways.
Choose the 4th digit either of the remaining 2 ways. 

That's 1×5×4×3×2 = 120 ways ending in 0

Case 2: 2 or 6 goes last:

Choose the last (5th) digit as in either of 2 ways.
Choose the 1st digit any of 4 ways. (CANNOT choose 0)
Choose the 2nd digit any of the remaining 4 ways. (CAN choose 0)
Choose the 3rd digit any of the remaining 3 ways.
Choose the 4th digit either of the remaining 2 ways.

That's 2×4×4×3×2 = 192 ways ending in 2 or 6

Total for the 2 cases = 120+192 = 312

Edwin