One classical query cannot solve Deutsch exactly

Fix a deterministic one-query procedure and let $q$ be its queried address. The constant-zero function $f_0$ and the balanced function $g$ defined by $g(q)=0$, $g(1-q)=1$ give the same reply. The procedure therefore has the same transcript and output on two functions requiring different answers, so it cannot be exact.

Querying both addresses reveals the entire truth table; equal replies identify a constant and unequal replies identify a balanced function. Thus two queries suffice exactly, and step 1.1 proves the deterministic exact query complexity is two.

Fix a random tape of an algorithm that makes at most one query. If it makes no query, its one output is correct on at most two of the four truth tables. If it queries address $q$, then for each possible reply $b\in \{0,1\}$ there are exactly two promised functions with $f(q)=b$: one constant and one balanced. The fixed tape gives the same output for both, so it is correct on at most one of each such pair, hence on at most two of the four functions. There are finitely many deterministic one-query strategies on two addresses with binary outputs. The random-tape algorithm induces a probability weight on this finite strategy set by grouping tapes according to their resulting strategy. Averaging the fixed-strategy bound over that finite distribution and over the four functions preserves the bound $1\mathrm{/}2$; by [F3] some promised input has success at most $1\mathrm{/}2$.