Regex for strings with even number of 1's and number of 0's divisible by 5
Your last idea is good, but insufficient. Also there is not the right number of zeros. Now let's try to decompose the problem:
Let's consider $L_0=\{1^i01^j01^k01^l01^m01^n,i+j+k+l+m+n \text{ is even}\}$.
Your (corrected) idea was $e_1 = (11)^*0(11)^*0(11)^*0(11)^*0(11)^*0(11)^*$, but this only generate words of $L_0$ where $i, j, k, l, m, n$ are all even. You need now to consider other possibilities. For example, if $j$ and $m$ are odd and the rest are even, we get the regular expression:
$$(11)^*0(11)^*10(11)^*0(11)^*0(11)^*10(11)^*$$
There are a total of $\binom{6}{0} + \binom{6}{2}+\binom{6}{4}+\binom{6}{6} = 32$ expressions to consider all possibilities (choice of an even number of indices that are odd among the 6).
The final regular expression will then be $\left(e_1+e_2+…+e_{32}\right)$.
Now you need to do the same thing with a language $L_1$ containg an odd number of $1$'s, and construct the regular expression for your language with both $L_0$ and $L_1$.
This sure is ugly as hell, but hey, it works! There may be a simpler way to write it, but I'm not so sure.