Here as U and V are uniformly distributed, they will have equal probabilities for each point in {0,1,2,3,4} which by the rule of probability comes $1/5$. Now for being uniformly distributed F(1), F(2), F(3) must have same values at points {0,1,2,3,4}
So,
$F(1) = (1+U +V) mod 5 = (1 + 1/5 + 1/5) mod 5 $ for all {0,1,2,3,4}
$F(2) = (1+2U +4V) mod 5 = (1 + 2/5 + 4/5) mod 5 $ for all {0,1,2,3,4}
$F(3) = (1+3U +9V) mod 5 = (1 + 3/5 + 9/5) mod 5 $ for all {0,1,2,3,4}
so all three are uniformly distributed as they all same value for all points.
but not identical as they have values different from each other.
hence Option C is false.