（1）∵tanα=2，∴原式= tanα+1 2tanα?1 = 2+1 4?1 =1；（2）∵tanα=2，∴原式= sin2α+sinαcosα+2cos2α sin2α+cos2α = tan2α+tanα+2 tan2α+1 = 4+2+2 4+1 = 8 5 ．
