Ytukyg

Let $L$ be a language over ${0,1}$ whose Nerode equivalence classes are:
$$
{w|#_0(w)mod2=0quadlandquad #_1(w)mod2=0}\
{w|#_0(w)mod2=0quadlandquad #_1(w)mod2=1}\
{w|#_0(w)mod2=1quadlandquad #_1(w)mod2=1}\
{w|#_0(w)mod2=1quadlandquad #_1(w)mod2=0}
$$
($#_0(w)mod2=0$ means that the number of zeroes in word $w$ is even).

Also $epsilon in L, 0,1,1110notin L$. Find DFA for the language.

First because there's a finite number of eq. classes by Nerode theorem we know that $L$ is regular so DFA exists for the language.

The solution to the problem is:

But I don't understand how the DFA accommodates all possible words in $L$.

1) For example what about $111$? It had even number of $0$'s and uneven number of $1$'s so it belongs to the second eq.class. But I don't see how we can arrive to this word using the DFA above.

2) What about the word $00011$? It belongs to the fourth equivalence class but I don't see how to arrive at this word using the DFA.

Not every equivalence class belongs to the language, otherwise the language would always be $Sigma^*$. Since $varepsilon in L$ and $0,1,1110 notin L$, we know that $L$ equals the first equivalence class. Therefore your examples are not counterexamples.