(a) The \(n\)th term is \(5730-23(n-1)\). It is negative if \[5730-23(n-1)<0\iff n>\frac{5753}{23}=250.1304\ldots.\] The least integer is \(251\), giving \[\boxed{u_{251}=5730-23(250)=-20}.\] The previous term is \(u_{250}=3\).

(b)(i) The common ratio is \[r_{\mathrm{GP}}=\frac{\tfrac12\tan\theta}{\sin\theta}=\frac1{2\cos\theta}>0.\] Convergence requires \(r_{\mathrm{GP}}<1\), so \(\cos\theta>1/2\). Within the given range, \(\boxed{0<\theta<\pi/3}\).

(b)(ii) At \(\theta=\pi/6\), the first term is \(1/2\) and the common ratio is \(1/\sqrt3\). Therefore \[S_\infty=\frac{1/2}{1-1/\sqrt3}=\frac{\sqrt3}{2(\sqrt3-1)}=\frac{\sqrt3(\sqrt3+1)}{2(3-1)}=\boxed{\frac{3+\sqrt3}{4}}.\] One choice is \(p=3,q=3,r=4\).