Minimisatiom using Myhill-Nerode Theorem
In the first automaton, you can see that every accepting state has a transition with 0 to an accepting state, while a transition with 1 goes to either a sink or is just non-existing. But this means that reading a 1 from an accepting state leads always to rejecting the input. Thus, all accepting states are equivalent.

In the second automaton, you can see that $q_2$ has a 0-transition to $q_5$, a non-accepting state, while the other two accepting states have 0-transitions to accepting states. Thus, $q_2$ cannot be equivalent to both $q_4$, and $q_3$. Moreover, $q_3$ and $q_4$ cannot be equivalent because reading 00 from both states leads to different behavior (one run is accepting, th other rejecting). Thus, in this case you cannot merge any of the accepting states.