 Solution

The gate set is universal. Since the Hadamard gate satisfies $H^{†}=H$ and $H^2=I$, the supplied $U$ is a unitary operator, andTwo 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 $(AS{)}^6=-iI$, henceThis finite product uses only $U$ and $T$, so the conclusion does not rely on discarding a global phase.