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For a planar Lebesgue measurable set $B$, its vertical section $B_x$ is a one-dimensional Lebesgue measurable set for $x$ almost everywhere. Write $B△C\subseteq Z$ with $C$ a planar Borel set and $Z$ a Borel Lebesgue null set. Every section $C_x$ and $Z_x$ is Borel; Tonelli's theorem gives $\int m(Z_x)dx=m_2(Z)=0$. Thus $m(Z_x)=0$ for $x$ almost everywhere, and $B_x△C_x\subseteq Z_x$ proves the assertion by Completeness of Lebesgue measure. No finite-measure condition on $B$ is needed.

Exceptional sections need not be measurable: $\{0\}\times A$ is a planar Lebesgue null set for every $A\subseteq \mathbb{R}$, even when $A$ is a non-measurable set. Conversely, if both $A_1$ and $A_2$ are non-measurable, $A_1\times A_2$ cannot be measurable. Otherwise every $x\in A_1$ would have to lie in the null exceptional set, making $A_1$ measurable by Completeness of Lebesgue measure.

False. Take $E=\{0\}$ and any non-measurable set $A\subseteq \mathbb{R}$. The vertical line $\{0\}\times \mathbb{R}$ is a planar Lebesgue null set: each segment $\{0\}\times [-N,N]$ can be enclosed in rectangles of arbitrarily small area, and the line is a countable union of these segments. Consequently its subset $E\times A$ is a Lebesgue measurable set of planar measure zero, by Completeness of Lebesgue measure.

A nonempty factor can therefore have too little Lebesgue measure to transmit the other factor's failure of measurability. Its exceptional vertical section at $x=0$ is consistent with Lebesgue measurable sections being guaranteed only almost everywhere.