= Solution
Write $n=2r$ and $k=2s$. Partition an $n$-element <set> into $r$ distinguished pairs, and let $\tau$ interchange the two elements of every pair. It induces an <involution> on the $k$-element subsets. Every nonfixed subset is paired with its distinct image, so nonfixed subsets contribute an even number to the <binomial coefficient> $\binom{2r}{2s}$.
A fixed subset contains either both or neither of the elements of each distinguished pair. Since it contains $2s$ elements, it consists of exactly $s$ whole pairs. There are $\binom rs$ such fixed subsets. Counting modulo $2$ therefore gives
$$
\boxed{\binom nk\equiv\binom{n/2}{k/2}\pmod2}.
$$
The same <involution> explains part (i): a fixed subset cannot have odd size.
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