Regex for strings with even number of 1's and number of 0's divisible by 5 I was trying to build up this regex but I am having huge expressions but nothing generates the given regex. I tried to break up the regex as

$0$ multiple of five times and odd $1$'s
$0$ multiple of five times and even $1$'s

Bu this doesn't help me.
The regex  I Found was  $(00)^*1(00)^*1(00)^*1(00)^*1(00)^*$. But in this way two $1$'s and five $0$'s cannot be generated.
Again $(11)^*0(11)^*0(11)^*0(11)^*0(11)^*$ does not generate $5n$ times $0$'s for $n\ge 2$.
There are other possibilities and that too do not meet the regex.
Any help would be appreciated.
P.s: I am at the very beginning in regex. So I am currently trying to avoid the method of conversion from finite automata to regular expressions.
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.