= Solution
\b[The gate set is universal.] Since the <Hadamard gate> satisfies $H^\dagger=H$ and $H^2=I$, the supplied $U$ is a <unitary operator>, and
$$
U^2=\frac{(I-iH)^2}{2}=-iH.
$$
Two applications of $U$ therefore realize a <Hadamard gate> up to a <global phase>. Together with the supplied <controlled-NOT gate> and <T gate>, this simulates the given <universal quantum gate set>.
An exact matrix representative of $H$ can also be synthesized if desired. Put $A=U^2=-iH$ and $S=T^2$. Direct multiplication gives $(AS)^6=-iI$, hence
$$
\boxed{(AS)^{18}A=H.}
$$
This finite product uses only $U$ and $T$, so the conclusion does not rely on discarding a <global phase>.
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