DFA for strings with number of 0's odd only in substring longer than 1
I believe your language is not regular and if I'm not mistaken, the pumping lemma suffices:
Let us assume by contradiction your language is regular, thus it should hold all 3 conditions of the pumping lemma.

Let $w=0^{2p+1}111$, notice $w\in L$ because $2n$ is always even, thus by adding $1$ we have an odd number of $0$s.

Now let us take a simple example where $p=1$, but this should work for any $p$. 

In this case, $p=1$, notice $|w|&gt;p$, since $|w|=5$, essentially $w=000111$. In our case this means $y=0$ and because $|xy|\le 1$ and because $|y|&gt;0$  where 1 is our pumping length this means $x=\epsilon$. 

This leaves $z=0011$. However, for $i=2$, we receive: 000011 which is $y^2z \equiv 000011 $.

Then the number of 0s in your substring is now even, meaning $000011 \notin L$. 
Again, we can do this for any pumping length $p$ by simply pumping up or pumping down.

Thus there won't be a DFA/NFA for that. You can try using a TM.

For more information: pumping lemma.