Whether can we infer that $X_1$ and $X_2$ are independent if we know they are conditional independence given $D$ and $\overline{D}$？ If we assume that:
(1) $P(X_1, X_2|D)=P(X_1|D)P(X_2|D)$, i.e., $X_1$ and $X_2$ are conditional independence given $D$;
(2) $P(X_1, X_2|\overline{D})=P(X_1|\overline{D})P(X_2|\overline{D})$, i.e., $X_1$ and $X_2$ are conditional independence given $\overline{D}$.
(PS: we cannot derive (2) with the assumption of (1))
Can we infer that $P(X_1,X_2)=P(X_1)P(X_2)$?
 A: Consider three binary variables $X_1, X_2, D$ taking values in $\{0, 1\}$. For $D=0$ let $(X_1, X_2) = (0, 0)$ with probability one, and for $D=1$ let $(X_1, X_2) = (1, 1)$ with probability one. Furthermore, let $D$ be uniform.
Then $X_1, X_2$ are independent conditioned on $D$, but they are clearly dependent (if you know that $X_1 = 1$ you also know that $X_2=1$, and the same for $X_1=0$ leading to $X_2=0$, both happening with positive probability).

Another possibility:
Take e.g. any pair $X_1, X_2$ of discrete dependent variables, and set $D=X_2$. Then $X_1, X_2$ are dependent but conditioned on $X_2$ they are independent:
$$
\begin{align}
P(X_1, X_2|X_2) &= P(X_1|X_2)\\
    &= P(X_1|X_2) \cdot 1\\
    &= P(X_1|X_2) \cdot P(X_2|X_2).
\end{align}
$$
A: No.
Here are confusion matrices for a counter-example:
given $D$:
[[1,2], [2,4]]
given $D'$:
[[6,3], [2,1]]
You can see that the odds of $X_1$ are 1:2 if $D$ and 2:1 if $D'$ independent of $X_2$.  If you add these two, though, you get:
[[7,5], [4,5]]
