(1)
（向量a-向量b）={-2cos x,2sin(x/2）-2cos(x/2)}
所以：│向量a-向量b│（即距离）=√{（2cos x)^2+[2sin(x/2）-2cos(x/2)]^2}=2√{(cos x)^2+[sin(x/2）-cos(x/2)]^2}=2√{(cos x)^2+[1-2sin(x/2）cos(x/2)]}=2√[(cos x)^2+(1-sin x)]
所以f(x)=2+sin x-(1/4)*│向量a-向量b│^2
=2+sin x-(cos x)^2-1+sin x
=1-(cos x)^2+2sin x
=(sin x)^2+2sin x
(2)
若函数f(x)和函数g(x)的图像关于原点对称,则点（x,y）关于原点的对称点为(-x,-y),设点（x,y）在g(x)上,则点(-x,-y)在f(x)上,所以：-y=[(sin-x)^2]+2sin-x
=(sin x)^2-2sin x(因为sin-x=-sin x)
所以y=-(sin x)^2+2sin x=g(x)
(3)
h(x)=g(x)-入f(x)+1
=-(1+λ)(sin x)^2+(2-2λ)sin x+1
=-(1+λ){(sin x)^2-[(2-2λ)/(1+λ)]sin x}+1
=-(1+λ){{sin x-[(1-λ)/(1+λ)]}^2-[(1-λ)/(1+λ)]^2}+1
因为h(x)在[-π/2,π/2]上是增函数,所以{sin x-[(1-λ)/(1+λ)]}^2在此区间上为减函数,所以sin x-[(1-λ)/(1+λ)]≤0因为sin x在[-π/2,π/2]的最大值为1,所以(1-λ)/(1+λ)≥1
解得-1＜λ≤0