解析：
f(x)=sin(2x-π/6)+cos² x
=sin2x*cos(π/6) - cos2x*sin(π/6) + (cos2x +1)/2
=sin2x*√3/2 - cos2x*1/2 + cos2x*1/2 +1/2
=(√3/2)*sin2x＋1/2
(1) 若f(a)=1,那么：
(√3/2)*sin2a＋1/2=1
(√3/2)*2sina*cosa=1/2
解得：sina*cosa=(√3)/6
(2)由上知：f(x)=(√3/2)*sin2x＋1/2
可知當(dāng)2kπ - π/2≤2x≤2kπ + π/2,即 kπ - π/4≤ x ≤kπ + π/4,k∈Z時(shí),函數(shù)f(x)是增函數(shù)
所以函數(shù)f(x)的單調(diào)增區(qū)間為[kπ - π/4,kπ + π/4],k∈Z