The zero point of g(x) = f(x)-k is the solution of f(x) = k, that is, the intersection of f(x)=|2sinx+m| and straight line y=k;

Let's look at the case of m=0, as shown in the figure.

The graph with f(x)=|2sinx+m| is two half waves with the same left and right. Obviously, the period is π, and the condition m=0 is enough.

Look at m>0, as shown.

The graph of f(x)=|2sinx+m| is two half waves with a period of 2π;

M<0 is similar to this, except that two and a half waves are left low and right high;

When m & gt=2 or m

Obviously, only when m=0, there will be two half waves with the same period π; In other cases, the period is 2π.

therefore

(1) is correct; (2) Correct

(3)(4)(5) It is necessary to investigate the intersection of y=f(x) and y = k.

According to the intersection of the above picture and the right picture in the following picture, the intersection may form a arithmetic progression:

a.？ There are two tangents in a period, which can be evenly distributed in each period to form a arithmetic progression.

b.？ A period has four intersections, and each period can be evenly distributed with four intersections, which can form a arithmetic progression.

c.？ There are two intersections and a tangent point in a period, and three points can be evenly distributed in each period to form a arithmetic progression.

In these three cases, the values of m and k are unique; And the tolerance of intersection coordinates is not greater than π;

D. when |m| > when =2, f(x) is a complete sine wave, and there is a case where k=|m|+2. The curves of y=k and y=f(x) have a tangent point in each period, and the tolerance between the tangent points is 2π;

But M, K satisfying this condition is not unique, there are countless (M, K) pairs, as long as? Satisfying K=|m|+2 can make the tolerance of intersection point 2π, as shown in the figure.

f.？ K=|m|, each period has two intersections, each period can be evenly distributed, and the coordinate tolerance is π.

Because the fifth question is incomplete, you can judge right or wrong by yourself according to the above analysis.