Hint for a):
Use the identity $\ln a−\ln b=\ln(a/b)$.
Hint for b):
What is $|\sin e^{\ln(2\pi x+\pi/2)}−\sin e^{\ln(2\pi x)}|$ and what does part a) tell you about $|\ln(2\pi x+\pi/2)−\ln(2\pi x)|$?
Warning: solution follows.
For part a), use the difference rules for logarithms: $ \ln(2\pi x+{\textstyle{\pi\over2}})-\ln(2\pi x)=\ln{ 2\pi x+{\pi\over2} \over 2\pi x }\quad \buildrel{x\rightarrow\infty}\over{\longrightarrow }\quad\ln (1)=0. $
For part b), note that, for $x$ an integer $ \bigl|\,\sin e^{ \ln(2\pi x+{\textstyle{\pi\over2}})} -\sin e^{\ln 2\pi x}\,\bigr| =\bigl |\,\sin (2\pi x+{\textstyle{\pi\over2}}) -\sin (2\pi x)\,\bigr |=1. $
Towards showing that $f(x)=\sin(e^x)$ is not uniformly continuous, let $\epsilon=1$ and suppose $\delta>0$.
By part a), we can select an integer $x$ so that $|\ln(2\pi x+{\textstyle{\pi\over2}})-\ln(2\pi x)|<\delta$.
But by part b), $\bigl|\,\sin e^{ \ln(2\pi x+{\pi\over2})} -\sin e^{\ln( 2\pi x)}\,\bigr|=1= \epsilon.$
So, we have demonstrated that for for $\epsilon=1$, and any $\delta>0$, we can find two points $x_1=\ln(2\pi x+{\pi\over2}) $ and $x_2=\ln(2\pi x)$ such that $|x_1-x_2|<\delta$, yet $\bigl|\,\sin e^{x_1} -\sin e^{x_2}\,\bigr|\ge\epsilon$. This shows that $\sin(e^x)$ is not uniformly continuous on $\Bbb R$.