Regularity of a language with odd and even positions containing an even number of 1's
To create the DFA, just start doing it. I assume there is no requirement that the number of elements in the string is even. 

You start in the state A. In state A, you have an even number of 1's in the even and odd positions (initially, both counts are 0), so A is an accepting state. To get to an accepting state again, there must be a string following with an even number of 1's in the even and odd positions. 

In state A, the next symbol can be 0 or 1. In case of 0, you have the exact same situation again, you need another string following with an even number of 1's in the even and odd positions. So A followed by 0 goes to A. 

If the next symbol is 1, then the rest of the string must have an even number of 1's in even positions, and an odd number of 1's in odd positions. That's another state B.

In state B, if the next symbol is 0 then you now need an even number of 1's in odd positions, and an odd number of 1's in even positions. That's a new state C. In state B, if the next symbol is 1 then you now need an odd number of 1's in even and odd positions. That's a new state D. 

Now you do the same for states C and D, and you find that whatever the input, you'll end up in one of the states A, B, C and D. 

Don't think about "state" as some abstract state in a DFA. Think of it as something that you can describe in words. From any state, which you can describe in words, you get to another state, which you can describe in words again. If you compare those descriptions, and they are the same, then the state is the same. 

The case where you don't get a DFA is when the number of states is not finite. For example, for $0^n1^n$, the initial state is "I need any number n of 0's and the same number n of 1's". If you get a 0 then you get a state where you need any number n of 0's and n+1 1's. Another 0 and you need n 0's and n+2 1's. And so on. The number of states is not finite. In your problem, there are only four states needed. 

If that's too practical, then given any prefix X of a string in the language L, define the set of suffixes Y such that XY is in the language. Any such set of suffices is a state.  An accepting state is one where the set contains the empty string. The initial state is the set of suffices when X is the empty string. In your case, there are only four different sets of suffices.