Undecidability and Unrecognizability of Language with two Turing Machines
You are going to want to use a many-one reduction for this problem. Note that $ii \implies i$ so it is sufficient to prove statement $ii$. We know that the complement to the halting problem $$ \bar{H}=\{\langle M, w \rangle \mid M \text{ does not halt on }w\}$$ is not recursively enumerable. We can reduce $\bar{H}$ to $Language1$
$S$ = &quot; 
$\quad$on input $\langle M,w \rangle$ 
$\quad$construct $M_1$ = &quot; 
$\quad\quad$on input $x$ 
$\quad\quad$if $x\neq $&lt;$M_2$&gt;, $Reject$ 
$\quad\quad$run $M$ on $w$ 
$\quad\quad$$Accept$ &quot; 
$\quad$construct $M_2$ = &quot; 
$\quad\quad$on input $x$ 
$\quad\quad$if $x\neq $&lt;$M_1$&gt;, $Reject$ 
$\quad\quad$run $M$ on $w$ 
$\quad\quad$$Accept$ &quot; 
$\quad$Output $\langle M_1, M_2 \rangle$ &quot;
Notice that if $M$ does not halt on $w$, then $M_1$ accepts string $\langle M_2 \rangle$ and $M_2$ accepts string $\langle M_1 \rangle$, and both reject all other strings, that is $L(M_1)=\{\langle M_2 \rangle \}$ and $L(M_2) = \{\langle M_1 \rangle \}$. On the other hand if $M$ does halt on $w$, $L(M_1)=L(M_2)=\emptyset$. Then $\bar{H} \leq_m Language1$, since $\bar{H}$ is not recursively enumerable, neither is $Language1$.