f(x)=2[√3cosxsinx+2(cosx)^2]-2[(sinx)^2+(2cosx)^2]-1
=√3sin2x-2(cosx)^2-3
=√3sin2x-cos2x-4
=2sin(2x-π/6)-4,
1.0<=x<5π/12,
∴-π/6<=2x-π/6<=2π/3,
sin(2x-π/6)∈[-1/2,1],
∴f(x)的值域是[-5,-2].
2.2x-π/6=π/2,x=π/3,a=(√3/2,1/2),b=(√3/2,1）,
|（1/√t)a+√tb|=|((√3/2）(1/√t+√t),1/(2√t)+√t)|
=√[(3/4)(1/√t+√t)^2+1/(4t)+1+t]
=√[(3/4)(1/t+2+t)+1/(4t)+1+t]
=√（1/t+7t/4+5/2),1/2<=t<=2,
设g(t)=1/t+7t/4+5/2,则
g'(t)=-1/t^2+7/4=[(7/4)t^2-1]/t^2=(7/4)[t+√(4/7)][t-√(4/7)]/t^2,
1/2<=t<√(4/7)时g'(t)<0,g(t)↓；t>√(4/7)时g(t)↑,
∴g(t)|min=g(√(4/7)=√7+5/2,g(1/2)=2+7/8+5/2=43/8,g(2)=1/2+7/2+5/2=9/2,
g(t)|max=9/2,
∴所求最小值=√（√7+5/2）,最大值=3√2/2.