A refinement of your first approach:
In numbers which end with $2$, the sum of the other two digits must be $4$ or $7$:
- $132$
the sum of the other two digits is 4
- $312$
the sum of the other two digits is 4
- $342$
the sum of the other two digits is 7
- $432$
the sum of the other two digits is 7
In numbers which end with $4$, the sum of the other two digits must be $5$ or $8$:
- $234$
the sum of the other two digits is 5
- $324$
the sum of the other two digits is 5
- $354$
the sum of the other two digits is 8
- $534$
the sum of the other two digits is 8
The condition that the number is only once divisible by $25$ means that it is a multiple of $25$ but not $25^2 = 625$. Since we are restricted to using the digits $2, 3, 4, 5, 6, 7$, the four-digit number will only be divisible by $25$ if the last two digits are $25$ or $75$. From these, we must remove those multiples of $625$ between $2000$ and $8000$ in which the only digits are $2, 3, 4, 5, 6, 7$. Those multiples are:
$$2500, 3125, 3750, \color{blue}{4325}, 5000, 5625, 6250, 6875, 7500$$
Assuming digits cannot be repeated, you correctly found the number of four-digit numbers that are divisible by $25$. However, $4325$ is also a multiple of $625$ in which the only digits used are in the set $\{2, 3, 4, 5, 6, 7\}$ and no digits are repeated. Therefore, there are $24 - 1 = 23$ numbers that satisfy the given condition.
Best Answer
Hint : Since all the digits have to be even, any possible number is divisible by $8$, if and only if the last two digits form a number divisible by $8$.