Is this language recognizable? Let $L = \{M: M\text{ halts on only one of 1100 or 0011 or 0011 or 1000}\}$. I'm trying to determine whether $L$ is decidable.  

I don't think it's even recognizable, but I'm not sure. Regardless, I think I can show it's undecidable. Suppose $L$ is decidable by machine $D$. Then we can construct machine $S$ that decides the accepting problem, i.e. $\mathrm{ACC} = \{(M,w): M\text{ accepts }w\}$, which is undecidable. 

For $S$ on input $(M,w)$, construct machine $M'$ with input $x$ such that if $x \neq 1100$, then loop, and if $x = 1100$, run $M$ on $w$. If $M$ accepts, then $M'$ halts. Next, run machine $D$ on $M'$. If $M'$ halts, then $D$ accepts. Otherwise, $D$ rejects. 

So, if $M$ accepts $w$, then $x=1100$ and $M'$ halts, so $D$ accepts. On the other hand, if $M$ does not accept $w$, then $M'$ loops (and does not accept anything), and $D$ rejects. But then $D$ decides $\mathrm{ACC}$, which is undecidable. Hence, $L$ is undecidable.

Am I on the right track?
To show that $L$&nbsp;is not recognizable, you can reduce the complement of the halting problem to it. $\overline{\mathrm{HALT}}$ is not recognizable because, if a language and its complement are both recognizable, then they're both decidable, and we know that $\mathrm{HALT}$ isn't recursive.

Given a Turing machine&nbsp;$M$ and input&nbsp;$w$, you want to accept iff $M(w)$ loops. So construct the machine&nbsp;$M'$ which halts if its input is $1100$ and simulates $M(w)$, otherwise. If $M'\in L$, then $M'$ cannot halt on $0011$ or $1000$,* which means that $M(w)$ doesn't halt.

* It seems there's a typo in the question, which mentions $0011$ twice. Obviously, this doesn't make any real difference.