(4) shows the difficulty of proving a finite presentation group is trivial or not. Moishe Cohen mensioned the first group is called Higman group. I can't see why the Higman group is infinite but the second one is trivial.
I guess for the second part there is an elementary combinatorics proof, and for the first part we need to deal with some classical groups as a surjective homomorphic image of the shown group. Both parts I don't know either now. Help!
I will give a few more details of my comment on proving the Higman group infinite, but you need to know a bit about HNN-extensions and amalgamated free products to understand it.
The group $G_2 := \langle a,b \mid a^b=a^2 \rangle$ is an HNN-extension of one infinite cyclic group by another, and so is infinite. In fact this is the Baumslag-Solitar group ${\rm BS}(1,2)$ and has been much-studied.
The group $G_3 := \langle a,b,c \mid a^b=a^2, b^c=b^2 \rangle$ is an HNN-extension of $G_2$ by $\langle c \rangle$, and is infinite. In fact it is easy to see from Britton's Lamma on HNN-extensyions that no nontrivial reduced word in the subgroup $\langle a,c \rangle$ is trivial in $G_3$, so that subgroup is free of rank $2$.
Now $\langle c,d, a \mid c^d=c^2, d^a=d^2 \rangle$ is isomorphic to $G_3$ and again the subgroup $\langle a,c \rangle$ is free.
So the Higman group $G_4 := \langle a,b,c,d \mid a^b=a^2, b^c=b^2, c^d=c^2, d^a=d^2 \rangle$ is a free product with amalgamation of the two copies of $G_3$ amalgmated on the subgroup $\langle a,c \rangle$. By the theory of free products with amalgamation, both subgroups embed into $G_4$, so it is infinite, and in fact the subgroups $\langle a,c \rangle$ and $\langle b,d \rangle$ are free.