Collision resistance of composed hash functions
Suppose you have found a collision for $H^2$, ie a pair $(x_1,x_2), x_1\neq x_2$ such that $H^2(x_1)=H=H^2(x_2)$.

Now let $F$ be a deterministic function (like $H^1$?). Does 
$$(F\circ H^2)(x_1)=F(H^2(x_1))=F(H)=F(H^2(x_2))=(F\circ H^2)(x_2)$$
yield anything interesting?