(N/A) To justify that the reaction is a redox reaction,we assign oxidation numbers to each atom:
$\overset{+1}{H_2} \overset{-2}{O} + \overset{0}{F_2}$ $\rightarrow \overset{+1}{H} \overset{-1}{F} + \overset{+1}{H} \overset{-2}{O} \overset{+1}{F}$
$1$. In this reaction,the oxidation number of $F$ in $F_2$ is $0$.
$2$. In $HF$,the oxidation number of $F$ is $-1$. Since the oxidation number decreases from $0$ to $-1$,$F_2$ is reduced.
$3$. In $HOF$,the oxidation number of $F$ is $+1$. Since the oxidation number increases from $0$ to $+1$,$F_2$ is oxidized.
$4$. Since both oxidation and reduction occur simultaneously,this is a redox reaction.