The fifth digit is $5$ as you said.
The second, fourth, sixth, and eighth digits are even, as you said.
The fourth digit must be $2$ or $6$ since the third digit is odd.
The fourth, fifth, and sixth digits must sum to a multiple of $3$, so the sixth digit must be $8$ or $4$ respectively (depending on whether the fourth digit is $2$ or $6$ respectively).
The 3-digit number formed by the sixth, seventh, and eighth digits must be divisible by $8$. Since this looks like "$8ab$" or "$4ab$" where $a$ is odd, the eighth digit must be $6$ or $2$, whichever is not used by the fourth digit.
So, the second digit must be $4$ or $8$, whichever is not used by the sixth digit.
So far, we have two choices:
$$x_1 4 x_3 2 5 8 x_7 6 x_9$$
$$y_1 8 y_3 6 5 4 y_7 2 y_9$$
We just need to put $1$, $3$, $7$, and $9$ in the blanks to make the first three digits add up to a multiple of $3$, and the last three digits add up to a multiple of $3$.
Let's try the first choice. The first three digits are either $147$ or $741$. In either case, the divisibility by $8$ forces the last three digits to be $963$. Finally, we check divisibility by $7$, to decide between $147$ or $741$ in the first three digits, but neither work.
So we are left with the second case. The divisibility by $8$ means the seventh digit $y_7$ is either $3$ or $7$. If it is $3$, we can try either $789$, $189$, $987$, or $981$ for the first three digits and see if the divisibility by $7$ holds; none of them work.
So we must have $y_7=7$, and we can try $183$, $189$, $381$ and $981$ for the first three digits; $381$ satisfies the divisibility by $7$. Thus the answer is