(a) \(PR=PS+RS=5a\). In the large right triangle, \[QR=5a\tan(\pi/6)=\frac{5a}{\sqrt3}.\] Pythagoras in \(RQS\) gives \[QS=\sqrt{a^2+\left(\frac{5a}{\sqrt3}\right)^2}=\boxed{a\sqrt{\frac{28}{3}}}.\] Thus \(c=28,d=3\).

(b)(i) Since \(RQS\) is right-angled and \(\angle RQS\) is acute, \[\sin(\angle RQS)=\frac{RS}{QS}=\frac1{\sqrt{28/3}}=\frac3{2\sqrt{21}}=\frac{\sqrt{21}}{14}.\] Therefore \(\angle RQS=\sin^{-1}(\sqrt{21}/14)\).

(b)(ii) The full angle at \(Q\) is \(\pi/2-\pi/6=\pi/3\). Subtract \(\angle RQS\): \[\boxed{\theta=\frac\pi3-\sin^{-1}\left(\frac{\sqrt{21}}{14}\right)}.\]