Answer
$$\eqalign{
& \left( {\text{a}} \right)\frac{1}{{{e^2}}} \cr
& \left( {\text{b}} \right)1 \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {{e^2} + \Delta x} \right) - 2}}{{\Delta x}} \cr
& {\text{Rewrite the limit}} \cr
& \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {{e^2} + \Delta x} \right) - 2}}{{\Delta x}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {{e^2} + \Delta x} \right) - \ln \left( {{e^2} + 0} \right)}}{{\Delta x - 0}} \cr
& {\text{The derivative of the function }}f\left( x \right){\text{ at the point }}x = a,{\text{ is}} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}},{\text{ then}} \cr
& \underbrace {\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {{e^2} + \Delta x} \right) - \ln \left( {{e^2} + 0} \right)}}{{\Delta x - 0}}}_{\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}}} \Rightarrow f\left( {\Delta x} \right) = \ln \left( {{e^2} + \Delta x} \right) \cr
& f'\left( x \right) = \frac{d}{{d\Delta x}}\left[ {\ln \left( {{e^2} + \Delta x} \right)} \right] = \frac{1}{{{e^2} + \Delta x}} \cr
& \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {{e^2} + \Delta x} \right) - \ln \left( {{e^2} + 0} \right)}}{{\Delta x - 0}} = f'\left( 0 \right) \cr
& f'\left( 0 \right) = \frac{1}{{{e^2} + 0}} = \frac{1}{{{e^2}}} \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\ln \left( {{e^2} + \Delta x} \right) - 2}}{{\Delta x}} = \frac{1}{{{e^2}}} \cr
& \cr
& \left( {\text{b}} \right)\mathop {\lim }\limits_{w \to 1} \frac{{\ln w}}{{w - 1}} \cr
& {\text{Rewrite the limit}} \cr
& \mathop {\lim }\limits_{w \to 1} \frac{{\ln w}}{{w - 1}} = \mathop {\lim }\limits_{w \to 1} \frac{{\ln w - \ln \left( 1 \right)}}{{w - 1}} \cr
& {\text{The derivative of the function }}f\left( x \right){\text{ at the point }}x = a,{\text{ is}} \cr
& f'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}},{\text{ then}} \cr
& \underbrace {\mathop {\lim }\limits_{w \to 1} \frac{{\ln w - \ln \left( 1 \right)}}{{w - 1}}}_{\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}}} \Rightarrow f\left( w \right) = \ln w \cr
& f'\left( w \right) = \frac{d}{{dw}}\left[ {\ln w} \right] = \frac{1}{w} \cr
& \mathop {\lim }\limits_{w \to 1} \frac{{\ln w - \ln \left( 1 \right)}}{{w - 1}} = f'\left( 1 \right) \cr
& f'\left( 1 \right) = \frac{1}{1} = 1 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{w \to 1} \frac{{\ln w}}{{w - 1}} = 1 \cr} $$