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