|a+b|=√(2+2cos2x)=√[2+2(2cos？ x- 1)]=√(4cos？ x)=2cosx

(2)f(x)=cos2x-2λ(2cosx)=2cos？ x-4λcosx- 1。

Let t=cosx, then f(x)=2t? -4λt- 1。

The symmetry axis is t=λ.

When λ≤ 1, t=- 1, f(x) is the smallest, and f(x)min=2+4λ- 1=-3/2, and λ=-5/4 is obtained.

When-1

When λ≥ 1, t= 1, f(x) is minimum, f(x)min=-4λ+ 1=-3/2, λ=-5/8 (give up).

So λ=-5/4.

The picture is too unclear. I did it according to your results.