1．已知$a = \text{e}^{- \frac{1}{2}}\sin\frac{1}{2}$，$b = \frac{1}{2}\text{e}^{- \frac{\text{π}}{6}}$，$c = \frac{\sqrt{2}}{2}\text{e}^{- \frac{\text{π}}{4}}$，则（ ）
A．$c > b > a$ B．$a > b > c$ C．$b > a > c$ D．$c > a > b$
解 令$f(x) = \text{e}^{x}\sin( - x) = - \text{e}^{x}\sin x$，则$f'(x) = - \text{e}^{x}\sin x - \text{e}^{x}\text{cos}x = - \sqrt{2}\text{e}^{x}\sin\left( x + \frac{\text{π}}{4} \right)$，
当$- \frac{\text{π}}{4} < x < 0$时，$\sin\left( x + \frac{\text{π}}{4} \right) > 0$，$\text{e}^{x} > 0$，则$f'(x) < 0$，所以$f(x)$在$\left( - \frac{\text{π}}{4},0 \right)$上单调递减，则$f(x) < f\left( - \frac{\text{π}}{4} \right)$，
因为$0 < \frac{1}{2} < \frac{\text{π}}{6} < \frac{\text{π}}{4}$，所以$0 > - \frac{1}{2} > - \frac{\text{π}}{6} > - \frac{\text{π}}{4}$，
所以$f\left( - \frac{1}{2} \right) < f\left( - \frac{\text{π}}{6} \right) < f\left( - \frac{\text{π}}{4} \right)$，
即$\text{e}^{- \frac{1}{2}}\sin\frac{1}{2} < \text{e}^{- \frac{\text{π}}{6}}\sin\frac{\text{π}}{6} < \text{e}^{- \frac{\text{π}}{4}}\sin\frac{\text{π}}{4}$，则$\text{e}^{- \frac{1}{2}}\sin\frac{1}{2} < \frac{1}{2}\text{e}^{- \frac{\text{π}}{6}} < \frac{\sqrt{2}}{2}\text{e}^{- \frac{\text{π}}{4}}$，故$a < b < c$，
所以$c > b > a$，故选：A.
2．设$a = \text{e}^{0.02} - 1$，$b = 2\left( \text{e}^{0.01} - 1 \right)$，$c = \sin 0.01 + \tan 0.01$，则（ ）
A．$a > b > c$ B．$a > c > b$
C．$c > a > b$ D．$b > c > a$
解 因为$a - b = \text{e}^{0.02} - 2\text{e}^{0.01} + 1 = \left( \text{e}^{0.01} - 1 \right)^{2} > 0$，所以$a > b$．
设$f(x) = 2\left( \text{e}^{x} - 1 \right) - \sin x - \tan x$，
则$f'(x) = 2\text{e}^{x} - \cos x - \frac{1}{\cos^{2}x}$，
令$g(x) = f'(x)$，则$g'(x) = 2\text{e}^{x} + \sin x - \frac{2\sin x}{\cos^{3}x}$．
当$x \in \left( 0,\frac{\text{π}}{6} \right)$时，$2\text{e}^{x} > 2$，$\sin x > 0$，$\frac{2\sin x}{\cos^{3}x} < \frac{2\sin\frac{\text{π}}{6}}{\cos^{3}\frac{\text{π}}{6}} = \frac{8\sqrt{3}}{9} < 2$，
所以$g'(x) > 0$，所以当$x \in \left( 0,\frac{\text{π}}{6} \right)$时，$f'(x) > f'(0) = 0$，
所以$f(x)$在$x \in \left( 0,\frac{\text{π}}{6} \right)$上单调递增，
从而$f(x) > f(0) = 0$，
因此$f(0.01) > 0$，即$b > c$．
综上可得$a > b > c$．故选：A