Yes, both distributive laws generalize. (You need only one of the two, but it’s useful to know both.) The first step in verifying the generalization that you want is to check that
$$\left(\bigcup_{i\in I}A_i\right)\cap D=\bigcup_{i\in I}(A_i\cap D)\;,\tag{1}$$
and to verify its mate you’ll want to check that
$$\left(\bigcap_{i\in I}A_i\right)\cup D=\bigcap_{i\in I}(A_i\cup D)\;.\tag{2}$$
Both are easily verified by element-chasing. For $(1)$, if $x\in\left(\bigcup_{i\in I}A_k\right)\cap D$, then $x\in\bigcup_{i\in I}A_i$ and $x\in D$. Since $x\in\bigcup_{i\in I}A_i$, there is an $i_0\in I$ such that $x\in A_{i_0}$, and therefore $x\in A_{i_0}\cap D\subseteq\bigcup_{i\in I}(A_i\cap D)$. Conversely, if $x\in\bigcup_{i\in I}(A_i\cap D)$, then there is an $i_0\in I$ such that $x\in A_{i_0}\cap D$. Then $x\in A_{i_0}\subseteq\bigcup_{i\in I}A_i$, and $x\in D$, so $x\in\left(\bigcup_{i\in I}A_i\right)\cap D$.
For $(2)$, if $x\in\left(\bigcap_{i\in I}A_i\right)\cup D$, then $x\in\bigcap_{i\in I}A_i$ or $x\in D$. Let $i_0\in I$ be arbitrary. Then $A_{i_0}\supseteq\bigcap_{i\in I}A_i$, so $x\in A_{i_0}$ or $x\in D$, and therefoer $x\in A_{i_0}\cup D$. Since this holds for each $i_0\in I$, $x\in\bigcap_{i\in I}(A_i\cup D)$. Conversely, if $x\in\bigcap_{i\in I}(A_i\cup D)$, then $x\in A_i\cup D$ for each $i\in I$. If $x\in D$, then certainly $x\in\left(\bigcap_{i\in I}A_i\right)\cup D$. If $x\notin D$, then we must have $x\in A_i$ for each $i\in I$, in which case $x\in\bigcap_{i\in I}A_i\subseteq\left(\bigcap_{i\in I}A_i\right)\cup D$.
Two applications of $(1)$ will give you the distributive law that you want. Suppose that $C=\bigcup_{i\in I}A_i$ and $D=\bigcup_{j\in J}B_j$. Then
$$\begin{align*}
C\cap D&=\left(\bigcup_{i\in I}A_i\right)\cap D\\
&=\bigcup_{i\in I}(A_i\cap D)\\
&=\bigcup_{i\in I}\left(A_i\cap\bigcup_{j\in J}B_j\right)\\
&=\bigcup_{i\in I}\left(\bigcup_{j\in J}(A_i\cap B_j)\right)\\
&=\bigcup_{\langle i,j\rangle\in I\times J}(A_i\cap B_j)\;.
\end{align*}$$
In other words, $C\cap D$ is the union of all possible intersections of the form $A_i\cap B_j$. In particular, if $I$ and $J$ are countable index sets, $I\times J$ is also countable.
Similarly, two applications of $(2)$ will give you the other general distributive law of this kind. This time suppose that $C=\bigcap_{i\in I}A_i$ and $D=\bigcap_{j\in J}B_j$. Then
$$\begin{align*}
C\cup D&=\left(\bigcap_{i\in I}A_i\right)\cup D\\
&=\bigcap_{i\in I}(A_i\cup D)\\
&=\bigcap_{i\in I}\left(A_i\cup\bigcap_{j\in J}B_j\right)\\
&=\bigcap_{i\in I}\left(\bigcap_{j\in J}(A_i\cup B_j)\right)\\
&=\bigcap_{\langle i,j\rangle\in I\times J}(A_i\cup B_j)\;
\end{align*}$$
the intersection of all possible unions of the form $A_i\cup B_j$ as $i$ and $j$ run over their respective index sets.
Yes. Prove, by induction on $n$, that any product of $n$ factors, no matter how it's parenthesized, agrees with the product of the same factors, in the same order, parenthesized "to the left", i.e., $((\dots((a_1a_2)a_3)\dots a_{n-2})a_{n-1})a_n$. You already know the first nontrivial case, $n=3$, so you only need to do the induction step.
Best Answer
If $\mathscr{A}$ is any family of sets, we can define
$$\bigcup\mathscr{A}=\{x:\exists A\in\mathscr{A}\,(x\in A)\}$$
and
$$\bigcap\mathscr{A}=\{x:\forall A\in\mathscr{A}\,(x\in A)\}\;.$$
These definitions specialize to the usual definition of $A\cup B$ and $A\cap B$ when $\mathscr{A}$ contains just two sets, and these definitions are independent of any parenthesizing or other organization of the members of $\mathscr{A}$ into smaller subcollections. Thus, the order and grouping of the sets makes no difference to the union and intersection of the family.