Counting $3$ digit even integers between $1$ and $1000$ with distinct digits $5$ choices for the last digit, $9$ choices for the second digit and $7$ choices for the first digit: $5 * 9 * 7$ integers with the given property. 
Or $5$ choices for the last digit, $8$ choices for the first digit and $8$ choices for the second digit: $5 * 8 * 8$ integers with the given property.
Since the results are not equal I am wondering which method is wrong and why. 
 A: In the first method: if your second digit is zero then you actually have eight choices for the first digit rather than seven. 
EDIT: As pointed out below, in fact both approaches are wrong: in the case the units digit is zero, then everything I said above goes out the window. So the actual correction for your first method is 
$$\underbrace{4 \cdot (\underbrace{8 \cdot 7}_{\text{Second digit } \neq 0} + \underbrace{8}_{\text{Second digit } = 0})}_{\text{Units digit is } 2,4,6,8}
+ \underbrace{9 \cdot 8}_{\text{Units digit is } 0} = 328.$$
The correction to your second method is
$$\underbrace{4 \cdot (8 \cdot 8)}_{\text{Units digit is } 2,4,6,8}
+ \underbrace{9 \cdot 8}_{\text{Units digit is } 0} = 328.$$
A: The best way to handle this problem is to divide it into two disjoint cases, depending on whether or not the units digit is zero.
Case 1:  If the units digit is zero, then there are nine choices for the hundreds digit (since we must exclude zero) and eight choices for the tens digit (since we must exclude both the hundreds digit and $0$). 
Case 2:  If the units digit is not zero, then there are four choices for the units digit ($2$, $4$, $6$, or $8$), eight choices for the hundreds digit (since we must exclude both $0$ and the units digit), and eight choices for the tens digit (since we must exclude both the hundreds digit and units digit).
Hence, the number of positive even integers with three distinct digits is 
$$9 \cdot 8 \cdot 1 + 8 \cdot 8 \cdot 4 = 72 + 256 = 328$$    
