Answer: \(z=166.8^\circ,\ 346.8^\circ\); \(\theta=-\dfrac{2\pi}{9},\ -\dfrac{\pi}{9},\ \dfrac{4\pi}{9}\).

For part (a),

\(2\operatorname{cot}(z+35^\circ)=5.\)

So

\(\operatorname{cot}(z+35^\circ)=\dfrac52.\)

Taking reciprocals,

\(\tan(z+35^\circ)=\dfrac25.\)

The reference angle is

\(\tan^{-1}\left(\dfrac25\right)=21.8^\circ.\)

Since \(0^\circ\leq z\leq360^\circ\), we have \(35^\circ\leq z+35^\circ\leq395^\circ\).

The valid values of \(z+35^\circ\) are

\(201.8^\circ\quad\text{and}\quad381.8^\circ.\)

Therefore

\(\boxed{z=166.8^\circ,\ 346.8^\circ}.\)

For part (b)(i), write everything in terms of \(\sin\theta\) and \(\cos\theta\):

\(\dfrac{\operatorname{sec}\theta}{\operatorname{cot}\theta+\tan\theta}=\dfrac{\dfrac1{\cos\theta}}{\dfrac{\cos\theta}{\sin\theta}+\dfrac{\sin\theta}{\cos\theta}}.\)

The denominator is

\(\dfrac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta}=\dfrac1{\sin\theta\cos\theta}.\)

Hence

\(\dfrac{\operatorname{sec}\theta}{\operatorname{cot}\theta+\tan\theta}=\dfrac1{\cos\theta}\cdot\sin\theta\cos\theta=\sin\theta.\)

Using the identity with angle \(3\theta\), the equation becomes

\(\sin3\theta=-\dfrac{\sqrt3}{2}.\)

Since \(-\dfrac{\pi}{2}\leq\theta\leq\dfrac{\pi}{2}\),

\(-\dfrac{3\pi}{2}\leq3\theta\leq\dfrac{3\pi}{2}.\)

In this interval,

\(3\theta=-\dfrac{2\pi}{3},\ -\dfrac{\pi}{3},\ \dfrac{4\pi}{3}.\)

Divide by \(3\):

\(\boxed{\theta=-\dfrac{2\pi}{9},\ -\dfrac{\pi}{9},\ \dfrac{4\pi}{9}}.\)