So the original function is a odd function, and the minimum positive period is π.

2. Function y= radical 3sinx+cosx=2 (radical 3/2 sinx+1/2 cosx) = 2 sin (x+π/6).

So the maximum value of the (1) function =2.

(2) use the function y= root number 3sinx+cosx(x∈R) to translate the image to the right, and the minimum distance can make the translated image symmetrical about the origin.

3. Function f(x)= 1-2sin? (x+π/8)+2sin(x+π/8)cos(x+π/8)

=cos(2x+π/4)+sin(2x+π/4)

= radical number 2[ radical number 2/2cos(2x+π/4)+ radical number 2/2sin(2x+π/4)]

= radical 2sin(2x+π/2)= radical 2cos(2x)

Therefore, the (1) function f (x) t = 2π/2 = the minimum positive period of π.

(2) The monotone decreasing interval 2kπ≤2x≤2kπ+π of the function f(x), that is, kπ≤ x≤ kπ+π, k ∈ r.

4. Because the vector m = (2cosx/2, 1), n = (sinx/2, 1) (x ∈ r), the function f(x)= m*n- 1.

Then f (x) = 2sinx/2cosx/2+1-1= sinx.

The range of (1) function f(x) is [- 1, 1].

(2) Because the three internal angles of acute angle △ABC are a, b, c, F (A) = 5/ 13, and F (B) = 3/5.

That is, sinA=5/ 13, sinB=3/5, COSA = 12/ 13, and COSB = 4/5 (acute angle △ABC).

Then f (c) = sinc = sin (a+b) = sinacosb+sinbcosa = 56/65.

5. Function f(x)=2cos? X+2 radical 3sinxcosx- 1 (x∈R)

=cos2x+ 1+ radical 3sin2x- 1=2sin(2x+π/6)

Then the period of the (1) function f(x) is 2π/2=π, and the symmetry axis equation is: x=π/6.

(2) Monotone decreasing interval of function f(x)

2kπ-π/2≤2x+π/6≤2kπ+π/2, that is, kπ-π/3≤x≤kπ+π/6, k ∈ R.

6. Function y=cosx-cos2x/2 (x∈R)

=cosx-cos？ x+ 1/2=-(cosx- 1/2)？ +3/4

The maximum value is equal to 3/4.

7. Function f(x)= radical number 3sinx+sin(π/2+x)= radical number 3sinx+cosx=2sin(x+π/6).

The maximum value is 2.