Let $F:\mathcal{A} \rightarrow \mathcal{B}$ be a right exact functor and suppose $\mathcal{A}$ has enough projectives.
Recall that the left hyper-derived functors of $F$ are defined as
$\mathbb{L}_{i}F(A_{*}) := H_{i}Tot(FP_{**})$
where $A_{*}$ is a chain complex in $\mathcal{A}$; $P_{**} \rightarrow A_{*}$ is a Cartan-Eilenberg resolution and Tot denotes the total complex of a double complex.
In Proposition 5.7.6, Weibel claims that there is a convergent spectral sequence
$E^{2}_{pq} = (L_{p}F)(H_{q}(A))\Rightarrow \mathbb{L}_{p+q}F(A) $ .
This comes from the second spectral sequence of a double complex, filtered by the rows.
I am fine with the RHS of the above, but for the left hand side, I think we have (using notation from Weibel )
$E^{2}_{pq} = H_{p}^{v}H_{q}^{h}(FP_{**}) = H_{p}^{v}(ker(FP_{qp} \rightarrow FP_{q-1 p})/ im(FP_{q+1 p} \rightarrow FP_{qp}))$.
To say that the previous line is equal to $(L_{p}F)(H_{q}(A))$, I think you need that
$H_{p}^{v}(ker(FP_{qp} \rightarrow FP_{q-1 p})/ im(F_{q+1 p} \rightarrow FP_{qp})) = H_{p}^{v}(F(ker(P_{qp} \rightarrow P_{q-1 p})/ im(P_{q+1 p} \rightarrow P_{qp})))$,
since $H_{q}(P,d^{h}) \rightarrow H_{q}(A)$ is a projective resolution given by the Cartan-Eilenberg resolution, so you should probably use it to compute the left derived functor.
If the above is true, I am not sure why $F$ commutes with homology, as we are only assuming it to be right exact.
I am probably being an idiot and missing something elementary. Any help would be much appreciated!