Collision resistance of composed hash functions Consider the following hash function

$H^{1\circ2}(x)=H^1(H^2(x))$

I have been asked to prove that if $H^1$ is CR,  $H^{1\circ2}$ will not generally itself be CR. I am assuming  $H^2$ is not CR, this is not explicitly stated. I have seen similar questions here but they always consider a double hash with a single function, not two.

I understand that by "not generally" or similar wording it is meant that there exists at least a case where $H^{1\circ2}$ is not CR. Admittedly however I thought that since $H^1$ is hashing $H^2(x)$, this makes the resulting hash function CR as well. Is this not always the case?
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?