= Solution
\b[The <Frankl-Wilson theorem>.] Let $p$ be a <prime number>, let $L\subseteq\mathbb F_p$ have $s$ elements, and let $\mathcal F\subseteq\mathcal P([n])$. Suppose that $|A|\bmod p\notin L$ for every $A\in\mathcal F$, whereas $|A\cap B|\bmod p\in L$ for all distinct members. Then
$$
\boxed{|\mathcal F|\leq\sum_{j=0}^{s}\binom nj,}
$$
with terms beyond $n$ understood as zero.
Work over the <finite field> $\mathbb F_p$. For each member define its <intersection polynomial>
$$
f_A(x_1,\ldots,x_n)=\prod_{\ell\in L}\left(\sum_{i\in A}x_i-\ell\right).
$$
Replace every positive power $x_i^d$ by $x_i$ to obtain a <multilinear polynomial> $\widetilde f_A$ of <polynomial degree> at most $s$. This <Boolean multilinearization> preserves its values at every <characteristic vector of a set>. At the <characteristic vector of a set> $B$,
$$
\widetilde f_A(\mathbf1_B)=\prod_{\ell\in L}(|A\cap B|-\ell).
$$
It is zero if $A\ne B$ and nonzero if $A=B$. Evaluating a proposed linear relation at each $\mathbf1_B$ therefore establishes <linear independence> of the $\widetilde f_A$. The <vector space> of <multilinear polynomials> of <polynomial degree> at most $s$ has the <basis> of <monomials> $\prod_{i\in T}x_i$, $|T|\leq s$, of size $\sum_{j=0}^{s}\binom nj$. This proves the <Frankl-Wilson theorem> by the <polynomial method in combinatorics>. The empty product for $s=0$ is $1$, and the argument still applies.
\b[The <constant-intersection family bound>, directly.] Write $\mathcal F=\{F_1,\ldots,F_q\}$ and assume $n\geq1$. The cases $q\leq1$ are immediate. If $\lambda=0$, the nonempty members are pairwise disjoint and there can also be the empty set, giving
$$
\boxed{q\leq n+1.}
$$
For $\lambda>0$, every member has size at least $\lambda$. If every member has size strictly greater than $\lambda$, form the real <matrix> $X$ whose rows are their <characteristic vectors of sets>. Its <Gram matrix> satisfies
$$
XX^{\mathsf T}=\operatorname{diag}(|F_1|-\lambda,\ldots,|F_q|-\lambda)+\lambda\mathbf1\mathbf1^{\mathsf T}.
$$
For any nonzero real <vector> $z$,
$$
z^{\mathsf T}XX^{\mathsf T}z
=\sum_i(|F_i|-\lambda)z_i^2+\lambda\left(\sum_i z_i\right)^2>0.
$$
Thus the rows are <linearly independent> and $q\leq n$. If a member $F_1$ has size $\lambda$, it lies inside every other member. The sets $F_i\setminus F_1$, $i>1$, are nonempty and pairwise disjoint, so $q-1\leq n-\lambda$. This also gives
$$
\boxed{q\leq n\quad(\lambda>0).}
$$
These bounds are sharp: the singletons together with $\varnothing$ attain $n+1$ at $\lambda=0$, while all complements of singletons attain $n$ at $\lambda=n-2>0$ when $n\geq3$.
\b[The <constant t-wise intersection dichotomy>.] In its substantive form the last argument requires $m\geq t$. The printed statement does not explicitly impose this. Without it the $t$-fold condition can be vacuous and the claimed dichotomy is false: take $t=3$, $m=2$, $n=1$, $A_1=\varnothing$, $A_2=\{1\}$ and $\lambda=1$. There is no $3$-tuple to test, no common $1$-set, and $k=0$, so $m=2>k+t-2=1$. We therefore prove the intended assertion for $m\geq t$ and record this necessary qualification.
If $\lambda=0$, the first alternative always holds with $S=\varnothing$. Suppose $\lambda>0$. Choose $t-2$ indices attaining the minimum intersection size $k$, and write
$$
K=\bigcap_{i\in I}A_i,\qquad |I|=t-2,\quad |K|=k.
$$
For each of the $q=m-t+2\geq2$ remaining indices put $B_j=K\cap A_j$. The hypothesis gives $|B_j\cap B_\ell|=\lambda$ for distinct remaining indices. This reduces the <constant t-wise intersection dichotomy> to a <constant-intersection family bound> on the $k$-element set $K$. The two possibilities are treated below.
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