Because f(x)=a+( 1-2a)/(x+2) is the increasing function in the interval (-2, positive infinity).

So 1-2a

So a> 1/2

So the range of a is (1/2, positive infinity).

1. 1)

f(x)=2cos？ x- 1+√3sin2x+α+ 1

=cos2x+√3sin2x+α+ 1

=2[sin(π/6+2x)]+ α+ 1

The minimum value of F(x) is a and the maximum value is 3+a in the range of [-π/6, π/6].

∴ 2a+3=3

a=0

2)f(x)=2[sin(π/6+2x)]+ 1

=2sin2(π/ 12+x)+ 1

Subtract the new graph from the old graph = vector (constant term moves left)

m=(π/ 12，- 1)

2.( 1) should be the range of finding k.

f(x)=cos？ kx-sin？ kx+2√3sinkxcoskx

=cos2kx+√3sin2kx

=sin(2kx+π/6)

From the original title:

ω≥π

2π/2k≥π

k≤ 1

(2) At this time, k= 1.

f(A)=sin(2A+π/6)= 1

A=π/6

From sine theorem:

2R=a/sinA=2√3

sinB+sinC=3/2R=√3/2

B+C=5π/6

Suppose c > b.

B=π/3，C=π/2

a=√3，b=3

S= 1/2 ab

=3√3/2