= Solution
{c}
Encode each point $x\in X$ by a partial <binary sequence> of length $m$: coordinate $i$ is fixed to zero if $x\in A_i$, fixed to one if $x\in B_i$, and unrestricted otherwise. Disjointness makes these prescriptions consistent. Let $d(x)$ be the number of fixed coordinates, equivalently the number of pairs containing $x$, and let $S_x\subseteq\{0,1\}^m$ be the set of complete sequences consistent with them. Then $|S_x|=2^{m-d(x)}$.
The <separating family of disjoint set pairs> ensures that $S_x$ and $S_y$ are disjoint whenever $x\ne y$: one coordinate prescribes opposite bits. Counting the sequences in these disjoint subcubes gives the <disjoint subcube packing inequality>
$$
\sum_{x\in X}2^{-d(x)}\leq1.
$$
Since $z\mapsto2^{-z}$ is a <convex function>, <Jensen inequality> implies
$$
n\,2^{-\bar d}\leq\sum_{x\in X}2^{-d(x)}\leq1,
\qquad
\bar d=\frac1n\sum_xd(x)=\frac1n\sum_i(|A_i|+|B_i|)\leq\lambda m.
$$
Taking logarithms yields $\log_2n\leq\bar d\leq\lambda m$. Therefore, for $\lambda>0$,
$$
\boxed{m\geq\left\lceil\frac{\log_2 n}{\lambda}\right\rceil.}
$$
The PDF omits the qualification $\lambda>0$. For $n\geq2$, it follows from feasibility: separation requires positive total incidence, so the stated incidence bound cannot hold with $\lambda\leq0$. For $n=1$, separation is vacuous, and $\lambda=0$ makes the printed quotient undefined; the intended parameter range is $0<\lambda\leq1$.
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