= Solution
\b[The statement is false.] Take two independent fair bits $U,V$ on the four equally probable outcomes $00,01,10,11$, and put $A_1=\{U=0\}$, $A_2=\{V=0\}$, and $B=\{U=V\}$. Each of $A_1,A_2,B$ has probability $1/2$, and
$$
P(B\cap A_1)=P(B\cap A_2)=\frac14=P(B)P(A_i).
$$
Thus $B$ is independent of each event separately. But $A_1\cup A_2$ contains $00,01,10$, whereas its intersection with $B$ contains only $00$. Therefore
$$
P(B\cap(A_1\cup A_2))=\frac14
\ne\frac12\cdot\frac34=P(B)P(A_1\cup A_2).
$$
This is a concrete failure of extending separate <independence> to the generated union. A counterexample at the permitted value $n=2$ disproves the universal claim; for any larger $n$ one can repeat $A_1$ as the additional events.
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