Clearly, $2019!!$ is an odd number and multiple of $125$. Therefore $2019!!$ can only end with $125$, $375$, $625$, or $875$. Their residues modulo $8$ are $5$, $7$, $1$, and $3$, respectively.

By the conclusion of # 4247, we know the product of any four consecutive odd number is congruent to $1$ modulo $8$. Because $2019!!$ is a product of $1010$ consecutive odd number and $1010\equiv 2\pmod{8}$. Therefore $$2019!!\equiv 1\times 3\equiv 3\pmod{8}$$

Hence the final answer is $\boxed{875}$.