(1) If the equation f(x)=0 has a real root.

Then discriminate >; =0

So m2-4 (m+1) (m-1) >; =0

m^2-4m^2+4>； =0

3m^2<； =4

m^2<； =4/3

-2√3/3 = & lt； m & lt=2√3/3

(2) If the inequality f(x) is greater than 0, the solution set is empty.

From f (x) > 0: (m+1) x 2-MX+m-1> 0.

That is, the minimum value of f(x) >; 0, that is, the images of the function are all above the X axis, and the opening of the function is upward.

Namely: (4ac-b2)/4a >; 0

m+ 1 & gt； 0，m & gt- 1 ....( 1)

Therefore: [4 (m+1) (m-1)-m2]/4 (m+1) > 0.

(3m^2-4)/4(m+ 1)>； 0

(3m^2-4)(m+ 1)>； 0

(m+2√3 /3)(m-2√3 /3)(m+ 1)>0

-2√3/3 & lt； M<- 1 or m & gt2√3 /3

Combination (1):

m & gt2√3 /3

(3) If the inequality f(x) is greater than 0, the solution set is r.

Discriminant > 0

(4ac-b^2)/4a>； 0

From (1), the value of m with discriminant greater than 0 is -2 √ 3/3.

Derived from (2), (4ac-b 2)/4a >; The value of m of 0 is m & gt2√3 /3.

Without such a value of m, the condition of (3) can be satisfied.