Turing recognizable & decidable: binary strings with even length. Let A = {(M) | M is a DFA such that L(M) is not the same as EVEN}
The language is clearly recognizable: given a DFA $M$, simulate it, in minlex order, on every word, and see if there is a word $w$ such that $|w|$ is odd, and $w$ is accepted by $M$, or if there exists a word $w$ such that $|w|$ is even, and $w$ is not accepted by $M$. In both cases - accept.

However, this machine doesn't show the decidability of the language.

In order to decide the language, it is easier to proceed as follows: Let $A$ be a DFA that accepts the language EVEN. Fix such a DFA.

Now, given $M$, we need to decide whether $L(M)=L(A)$. If you feel comfortable with automata, you can just say that this is decidable, and that's it. 

If you're not that comfortable yet, consider the following:
Given two DFAs $B,C$, it is easy to construct a DFA $D$ such that $L(D)=L(B)\cap \overline{L(C)}$, and it is also easy (well, decidable in NLOGSPACE) whether $L(D)=\emptyset$. The latter test is equivalent to deciding whether $L(B)\subseteq L(C)$. Thus, you can decide whether $L(M)\subseteq L(A)$ and $L(A)\subseteq L(M)$, which amounts to deciding whether $L(A)=L(M)$, and you are done.