Is { ww' | HamDist(w,w')>1 } context-free?
The intersection with $R=\{0^*10^*10^*10^*\mid$ even length words$\}$ is context-free, because a PDA can remember two positions, in a way. Anyhow, let's first see what this language $L$ is. Its complement is
$R\setminus L=\{0^a10^b10^c10^d\mid b=n/2 \vee c=n/2 \vee a+d=n/2\}$. Therefore,
$L=\{0^a10^b10^c10^d\mid b\ne n/2 \wedge  c\ne n/2 \wedge a+d\ne n/2\}$. We can rewrite this as
$L=\{0^a10^b10^c10^d\mid b&gt; n/2 \vee  c&gt; n/2 \vee a+d&gt; n/2 \vee b,c,a+d&lt;n/2\}$.

The first $3$ cases can be easily verified, and so can the fourth one.

$b&gt;n/2$: Start putting in the stack until the first 1, then start popping from the stack until it's non-empty. After it's empty, again start putting in the stack until we reach the second 1. From then on pop the stack.

$c&gt;n/2$: Same.

$a+d&gt;n/2$: Start putting in the stack until the first 1, then start popping from the stack until it's non-empty. After it's empty, again start putting in the stack until we reach the third 1. From then on pop the stack.

$b,c,a+d&lt;n/2$: Start putting in the stack until the first 1, then start popping from the stack until it's non-empty. After it's empty, again start putting in the stack until we reach $a+n/2$ (guessed non-deterministically, somewhere between second and third 1). From then on pop the stack.