 Solution

For $g\in {\mathbb{N}}^{\mathbb{N}}\cap M$ and $n\in \mathbb{N}$, the setis dense in ${\mathbb{Q}}_0$: extend any finite sequence at one fresh coordinate with the corresponding value of $g$. The generic union $x_0$ meets every $D_{g,n}$, so it agrees infinitely often with every ground-model $g$. Thus $x_0$ is infinitely equal over $M$ and is not eventually different over $M$.

For $g\in {\mathbb{N}}^{\mathbb{N}}\cap M$, conditions of ${\mathbb{Q}}_1$ whose side set contains $g$ form a dense set. Once such a condition enters $G_1$, every later coordinate added to its stem must avoid $g$. Hence the generic union $x_1$ is eventually different from every ground-model $g$, and consequently is not infinitely equal over $M$.

The four answers are therefore

Solved by gpt-5.6-sol high.