Answer
$$\eqalign{
& \left. a \right){\text{ }}f\left( x \right) = \cot x,{\text{ }}a = \frac{\pi }{4} \cr
& \left. b \right){\text{ }} - 2 \cr} $$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \pi /4} \frac{{\cot x - 1}}{{x - \frac{\pi }{4}}} \cr
& or \cr
& \mathop {\lim }\limits_{x \to \pi /4} \frac{{\cot x - \cot \left( {\pi /4} \right)}}{{x - \pi /4}} \cr
& {\text{The derivative of a continuous }}f\left( x \right){\text{ at the point }}a{\text{ is given by}} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}} \cr
& {\text{Then comparing both expressions we have,}} \cr
& \underbrace {\mathop {\lim }\limits_{x \to \pi /4} \frac{{\cot x - 1}}{{x - \frac{\pi }{4}}}}_{\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}}} \Rightarrow f\left( x \right) = \cot x{\text{ and }}a = \frac{\pi }{4} \cr
& {\text{Therefore,}} \cr
& \left. a \right){\text{ One possible }}f\left( x \right){\text{ and }}a{\text{ is:}} \cr
& f\left( x \right) = \cot x,{\text{ }}a = \frac{\pi }{4} \cr
& \cr
& \left. b \right){\text{ Evaluating the limit}} \cr
& \mathop {\lim }\limits_{x \to \pi /4} \frac{{\cot x - 1}}{{x - \frac{\pi }{4}}} \Rightarrow f'\left( x \right) = \frac{d}{{dx}}\left[ {\cot x} \right] \cr
& {\text{ }} \Rightarrow f'\left( x \right) = - {\csc ^2}x \cr
& {\text{Evaluating at }}a = \frac{\pi }{4} \cr
& f'\left( {\frac{\pi }{4}} \right) = - {\csc ^2}\left( {\frac{\pi }{4}} \right) = - 2 \cr
& {\text{,Then}} \cr
& \mathop {\lim }\limits_{h \to 0} \frac{{\cos \left( {\frac{\pi }{6} + h} \right) - \frac{{\sqrt 3 }}{2}}}{h} = - \frac{1}{2} \cr} $$
