= Solution
Write the normalized <qubit> as $|\phi\rangle=c_0|0\rangle+c_1|1\rangle$, with $|c_0|^2+|c_1|^2=1$. Only the $|1\rangle$ amplitude changes between the two <phase gates>, so
$$
\bigl(R(\theta+\delta)-R(\theta)\bigr)|\phi\rangle
=c_1e^{i\theta}(e^{i\delta}-1)|1\rangle.
$$
Since $|e^{i\delta}-1|=2|\sin(\delta/2)|\leq|\delta|$,
$$
\boxed{\|R_m|\phi\rangle-R(\theta)|\phi\rangle\|
=2|c_1||\sin(\delta/2)|\leq|\delta|.}
$$
The <phase-gate discretization error> is uniform over all input states; its exact <operator norm> is $2|\sin(\delta/2)|$. To choose an approximating gate, round $N\theta/(2\pi)$ to the nearest integer and reduce it modulo $N$. Taking the shortest circular angle difference gives $|\delta|\leq\pi/N$, and hence a uniform state-vector error at most $\pi/2^n$. The circular choice handles the identified endpoints $0$ and $2\pi$ correctly.
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