Answer
(a) $a=\pi/3,\ \ b=\pi/6$
(b) $a= \pi$
(other answers are possible.)
Work Step by Step
(a) Consider the values: $a=\pi/3,\ \ b=\pi/6$
Then we have
\begin{align*}
\cos a&= \frac{1}{2}\\
\cos b&= \frac{\sqrt{3}}{2}
\end{align*}
But $$\cos (a+b)=\cos (\pi/2)=0$$
Thus, we see that
$$\cos (a+b) \neq \cos a+\cos b$$
(b) Consider $a= \pi$:
\begin{align*}
\cos(a/2)&=0\\
\frac{\cos a}{2}&=-0.5
\end{align*}
Thus $$\cos \frac{a}{2} \neq \frac{\cos a}{2}$$
