= Solution
For $N\geq2$, induction using the <Grover rotation angle> gives
$$
G^k|\psi_A\rangle=\sin((2k+1)\theta)|a\rangle+\cos((2k+1)\theta)|\omega\rangle.
$$
The <quantum measurement in the computational basis> therefore returns $a$ with <probability> $\sin^2((2k+1)\theta)$. Since $0<\theta\leq\pi/4$, the first angle interval attaining the required success is
$$
\frac\pi2-\theta\ \leq\ (2k+1)\theta\ \leq\ \frac\pi2+\theta.
$$
The <first successful Grover iterate> is consequently
$$
\boxed{k_{\min}=\left\lceil\frac{\pi}{4\theta}-1\right\rceil
=\left\lceil\frac{\pi}{4\theta}\right\rceil-1.}
$$
Indeed, this is the first integer at or above the lower endpoint requirement $k\geq\pi/(4\theta)-1$. The inequality $\lceil t-1\rceil<t$ also implies $(2k_{\min}+1)\theta<\pi/2+\theta$, so this iterate cannot overshoot the first success interval. Every smaller nonnegative integer lies before that interval and fails the threshold. Finally, $\arcsin u>u$ for $0<u<1$, so
$$
\boxed{k_{\min}<\frac{\pi}{4\theta}<\frac{\pi\sqrt N}{4}.}
$$
For the singleton case $N=1$, measuring the initial state already returns $a$ with certainty, and $k_{\min}=0<\pi/4$. For $N=2$, the requested threshold is only $1/2$, so zero iterations already suffice. These boundary cases agree with the search guarantee.
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