a)Xét: $f(x)=\sin x+\tan x-2x,x\in (0,\frac{\pi}{2})$
$f'(x)=\cos x+\frac{1}{\cos x^{2}}-2>2\sqrt{\frac{1}{cos x}}-2>2-2=0$
$\Rightarrow f'(x)>0 \Rightarrow f(x)>f(0)=0$
Vậy: $\sin x+\tan x>2x,$ với $\forall x \in (0,\frac{\pi}{2})$
b)Áp dụng BĐT Cauchy:
$2^{\sin x}+ 2^{ \tan x}\geq 2\sqrt{2^{\sin x+\tan x}}\geq 2.2^{\frac{1}{2}(\sin x+\tan x)}>2.2^{x}$
vậy:$2^{\sin x}+ 2^{ \tan x} > 2^{ x+1} $ , với $\forall x \in (0,\frac{\pi}{2})$
$\Rightarrow$ (ĐPCM)