Unambiguous "weak" vector valued $L^{+\infty}$ spaces?

Question

For some time, I have been stuck to the problem to be described as follows. The (perhaps not so commonly known) facts given here are taken from R. E. Edwards' Functional Analysis (Holt, Rinehart and Winston 1965) pp. 578−589. Let $I=[0,1]$ with the Lebesgue measure, and consider real separable (not necessarily reflexive) Banach spaces $E$ with strong dual $F=E'_\beta$. A certain kind of "$L^{+\infty}(I,F)$", $\Lambda(I,E')$, representing the dual of $L^1(I,E)$ is constructed as follows. Let $Y$ be the vector space of a.e. bounded (i.e. bounded outside some set of measure zero) functions $g:I\to F$ such that ${\rm ev}_x\circ g:I\to\mathbb R$ given by $t\mapsto(g(t))(x)$ is measurable for all $x\in E$. Letting $N_0$ be the subspace formed by $g\in Y$ vanishing a.e., then $\Lambda(I,E')=Y/N_0$ becomes a Banach space when we equip it with the essential supremum norm of representatives $g$ of the equivalence classes $[g]=\{g+h:h\in N_0\}$. Then a linear homeomorphism $\Lambda(I,E')\to(L^1(I,E))'_\beta$ is given by $[g]\mapsto\ell:[f]\mapsto\int_I(g\ .f)$ where $(g\ .f)(t)=(g(t))(f(t))$.

The problem is now the following. Since generally preduals are not unique, there may be different (separable) spaces $E$ having linearly homeomorphic duals $F$. So, at least a priori, we cannot invariantly define some space "$\Lambda(I,F)$" as a certain kind of "$L^{+\infty}(I,F)$". According to this Philip Brooker's answer, there are nonisomorphic separable spaces $C(S)$ having isomorphic duals. One may then ask, whether (1) the corresponding spaces $\Lambda(I,(C(S))')$ are (isometrically) isomorpic or linearly homeomorphic, under the associated "natural" maps. Further, the dual of $L^1(I\times I)$ is represented by $L^{+\infty}(I\times I)$. Since $L^1(I\times I)$ is isomorphic to $L^1(I,L^1(I))$, we see that $L^{+\infty}(I\times I)$ is isomorphic to $\Lambda(I,(L^1(I))')$. One may then ask, whether (2) there are separable Banach spaces $E$ not linearly homeomorphic to $L^1(I)$, but having dual linearly homeomorphic to $L^{+\infty}(I)$ and $\Lambda(I,E')$ not linearly homeomorphic to $L^{+\infty}(I\times I)$ under the associated natural maps.

So, there are two concrete questions (1) and (2) above.

As an explanation of the phrase "natural map" above, I add the following. If $\ell_0$ is a linear homeomorphism $(C(S_1))'_\beta\to(C(S_2))'_\beta$, then the question is about whether a linear homeomorphism $\Lambda(I,(C(S_1))')\to\Lambda(I,(C(S_2))')$ is given by $[g]\mapsto[\ell_0\circ g]$. For the second question, if $\ell_0$ is a linear homeomorphism $E'_\beta\to L^{+\infty}(I)$, then the question is about whether a linear homeomorphism $\Lambda(I,E')\to L^{+\infty}(I\times I)$ is given by $[g]\mapsto[\hat g]$ where $[\hat g(t,\cdot)]=\ell_0(g(t))$ for suitably chosen $\hat g$.

I have above taken the attitude that the isomophism (or linear homeomorphism) class the space $\Lambda(I,E')$ is not solely determined by that of $E'_\beta$ but depends also on $E$. If someone knows the contrary to be true, I am gratefull for a reference or a proof. Also possible couterexamples where for separable Banach spaces $E,E_1$ with $E'_\beta$ and $(E_1)'_\beta$ linearly homeomorphic but $\Lambda(I,E')$ and $\Lambda(I,E_1')$ not, where the spaces $E,E_1$ are not some $C(S)$ or $L^1(I)$ as I suggested above, are wellcome.

Edited. (25.5.2013) The question (1) above is settled since in the case where $E=C(S)$ with $S$ a countable ordinal with the order topology has dual isomorphic to $\ell^{\\\,1}(\mathbb N_0)$ using Pettis' theorem and the dominated convergence theorem one can show that measurability of $I\owns t\mapsto g(t)(x)\in\mathbb R$ for all $x\in E$ implies (strong) measurability into $E_\beta^{\prime}\\,$.