Suppose the number begins with $3$. Then it must end with a digit small enough so that the product (which would be the middle digit) is $<10$. That admits $1+[9/3]=4$ choices where $[x]$ is the greatest integer less than or equal to $x$ and the possibility of a zero units digit is included. Do this for all nine possible initial digits and you get
$9+[9/1]+[9/2]+[9/3]+...+[9/9]=32$
Case $1$: The number is a single digit.
In this case, the only even numbers are $2$ and $4$, giving a total of $2$.
Case $2$: The number has exactly two digits.
In this case, the last digit must be either $2$ or $4$, and the first digit must be one of the other four digits allowed, giving a total of $2 \cdot 4=8$.
Case $3$: The number has exactly three digits.
In this case, the last digit must be either $2$ or $4$ and the middle digit must be one of the other four digits allowed.
If the middle digit is $5$, then the first digit must be one of the other three digits allowed, giving a total of $2 \cdot 3=6$.
If the middle digit is not $5$, then the first digit must be one of two digits other than the last two digits and $5$, giving a total of $2 \cdot 3 \cdot 2=12$.
So, there are $2+8+6+12=28$ even numbers less than $500$ with distinct digits $\in \{1,2,3,4,5\}$.
Best Answer
Let $a_n$ be the number of $n$-digit numbers with digits from $\{2,3,4\}$ with an odd number of $3$s. Consider two mutually exclusive cases:
These two cases imply that $$a_n = 2a_{n-1} + (3^{n-1}-a_{n-1}) = a_{n-1} + 3^{n-1}.$$ The initial condition is $a_0=0$, and iterating the recurrence yields $$a_n = \sum_{k=0}^{n-1} 3^k = \frac{3^n-1}{3-1} = \frac{3^n-1}{2}.$$