解；（1）f(x)＝cos2x+

3

sinxcosx+1=

1

2

cos2x+

3

2

sin2x+

3

2

=sin（2x+

π

6

）+

3

2

函数的周期T=

2π

2

=π
∵-1≤sin（2x+

π

6

）≤1
∴

1

2

≤sin（2x+

π

6

）+

3

2

≤

5

2

即

1

2

≤f（x）≤

5

2

（2）当-

π

2

+2kπ≤2x+

π

6

≤

π

2

+2kπ⇒x∈[-

π

3

+kπ，

π

6

+kπ]为函数的单调增区间．