= Solution
Let $R=\operatorname{diag}(1,e^{i\pi/12})$. Since the <Hadamard gate> maps the standard <basis> to $|+\rangle,|-\rangle$, the given gate is $Q=HRH$. Because $H^2=I$,
$$
Q^3=HR^3H=HTH,\qquad\boxed{T=HQ^3H.}
$$
Next, use $HZH=X$ on the target wire of the <Controlled-Z gate>:
$$
\boxed{\mathrm{CNOT}=(I\otimes H)\,CZ\,(I\otimes H).}
$$
The control-zero block remains the identity, and the control-one block becomes $HZH=X$, proving the identity with the stated control convention. We have therefore constructed each gate in the assumed <universal quantum gate set> $\{\mathrm{CNOT},H,T\}$ using only $CZ,Q,H$.
Replacing every gate of a circuit by its displayed finite construction reproduces that circuit exactly. Thus \b[$\{CZ,Q,H\}$ is universal] under the usual arbitrary-accuracy convention: it can perform every approximation obtainable from the given universal set. The claim concerns approximation of arbitrary <unitary operators>; it does not assert that finitely many circuits exactly represent an uncountable continuum of gates. This is <universality transfer by exact gate synthesis>.
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