Hint: There are two types of $4$ digit odd integers with all digits different: (i) First digit is even ($2,4,6,8$) or (ii) first digit is odd. Count the Type (i) numbers, the Type (ii) numbers, and add.
Type (i): There are $4$ choices for the first digit. For each of these choices there are $5$ choices for the last digit. For each of the choices we have made so far, there are $8$ choices for the second digit, and for each such choice $7$ choices for the third, a total of $(4)(5)(8)(7)$.
Now it's your turn: make a similar calculation for Type (ii).
The analysis for the even numbers will be roughly similar. The two cases are (i) last digit is $0$ and (ii) last digit is not $0$.
Remark: The tricky thing is that the first digit cannot be $0$. A slightly nicer approach, I think, is to first allow $0$ as a first digit, and then take away the "numbers" with first digit $0$, which we should not have counted.
So for odd numbers, if we allow $0$ as a first digit, we can choose the last digit in $5$ ways, and then for the rest we have $(9)(8)(7)$ choices, for a total of $(5)(9)(8)(7)$.
Now if we have $0$ as the first digit, the rest can be filled in $(5)(8)(7)$ ways.
So our answer is $(5)(9)(8)(7)-(5)(8)(7)=2240$.
For the evens, the same analysis gives $(5)(9)(8)(7)-(4)(8)(7)$.
Best Answer
For the case in question, if there is exactly one $7$ and two $3$'s, the other digit must be a $1$ or $8$. We have two ways of choosing this number. We have $\binom{4}{2}$ ways of choosing the positions of the two $3$'s, which leaves two ways to choose the position of the $7$, and one way to place the remaining digit.
$$2 \cdot \binom{4}{2} \cdot 2 \cdot 1 = 2 \cdot \frac{4!}{2!2!} \cdot 2! = 2 \cdot \frac{4!}{2!}$$ The factor of $2!$ in the denominator represents the number of ways we could permute the two $3$'s within a given arrangement without producing an arrangement distinguishable from that arrangement.