Grothendieck needs only left exactness

Set $F=G={\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Z}\mathrm{/}2,-)$ and $A=\mathbb{Z}$. Both functors are additive and left exact, since maps into a kernel are precisely maps killed by the next arrow. Evaluation at $1$ identifies $F(B)$ with $B[2]$, so $GF(B)=(B[2])[2]=B[2]=F(B)$ naturally. The complexes $0\to \mathbb{Z}\to \mathbb{Q}\to \mathbb{Q}\mathrm{/}\mathbb{Z}\to 0$ and $0\to \mathbb{Z}\mathrm{/}2\to \mathbb{Q}\mathrm{/}\mathbb{Z}\stackrel{2}{\to }\mathbb{Q}\mathrm{/}\mathbb{Z}\to 0$ are injective resolutions by F3: rational division proves divisibility, and the displayed kernels and images prove exactness.

The rank-one free resolution $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}2\to 0$ computes ${\mathrm{Ext}}^j(\mathbb{Z}\mathrm{/}2,B)$ as the cohomology of $B\stackrel{2}{\to }B$, by F2. Hence $R^{q}F(\mathbb{Z})$ is $\mathbb{Z}\mathrm{/}2$ at $q=1$ and zero elsewhere, while $R^{p}G(\mathbb{Z}\mathrm{/}2)$ is $\mathbb{Z}\mathrm{/}2$ at $p=0,1$ and zero elsewhere. The proposed $E_2$ therefore has precisely two nonzero entries, at $(0,1)$ and $(1,1)$. Every $d_r$ for $r\ge 2$ has zero source or target, so both survive. In total degree two the finite filtration would force $R^2(GF)(\mathbb{Z})\cong \mathbb{Z}\mathrm{/}2$. But $GF=F$ and the same length-one calculation gives $R^{2}F(\mathbb{Z})=0$, a contradiction.

The missing hypothesis really fails: $I=\mathbb{Q}\mathrm{/}\mathbb{Z}$ is injective, $F(I)=I[2]\cong \mathbb{Z}\mathrm{/}2$, and $R^{1}G(F(I))\cong \mathbb{Z}\mathrm{/}2\mathrm{\ne }0$. This is exactly the acyclicity used in F1's target comparison. Thus the counterexample retains explicit resolution existence and isolates the failure of injective-image acyclicity. AC was used only to license divisibility as injectivity; the displayed finite Ext calculations and failure of the conclusion are algebraic.