解: (1).f(x)=sinx+(1-cosx)/2-(1+cosx)/2+a.
=sinx+cosx+a.
∴fx)=√2sin(x-π/4)+a.
當(dāng)sin(x-π/4)=-1時(shí),f(x)min=√2.
√2*(-1)+a=√2.
∴a=2√2.
(2).∵ sinx的x∈(2kπ-π/2,2kπ,+ρ/2)為增函數(shù), x∈(2kπ+π/2,2kπ+3π/2)為減函數(shù).
f(x)=sin(x-π/4)+2√2的單調(diào)遞增區(qū)間為:x∈(2kπ-π/4,2kπ+π/4);
f(x)的單調(diào)遞減區(qū)間為:x∈(2kπ+π/4,2kπ+5π/4),k∈Z.