what could be the answer and how to solve it....?

In expansion, of [2x - (1/2x)]^6, the general term
= C(6,r) x^(6 - r) (1/2x)^r
= C(6,r) (1/2)^r x^(6 - r) x^-r
= C(6,r) * (1/2)^r x^(6 - 2r)

For the constant term in expansion of [2x - (1/2x)]^6 :
x^(6 - 2r) = x^0
6 - 2r = 0
r = 3
Then, the constant term in expansion of [2x - (1/2x)]^6
= C(6, 3) * (1/2)^3 x^(6 - 2*3)
= 20 * (1/8)
= 5/2

For the ^2 term in expansion of [2x - (1/2x)]^6 :
x^(6 - 2r) = x^2
6 - 2r = 2
r = 2
Then, the x^2 term in expansion of [2x - (1/2x)]^6
= C(6, 2) * (1/2)^2 x^(6 - 2*2)
= 15 * (1/4) x^2
= (15/4) x^2

Hence, [2x - (1/2x)]^6 = (5/2) + (15/4) x^2 + ……

The expansion of (1 + x^2) [2x - (1/2x)]^6
= (1 + x^2) [(5/2) + (15/4) x^2 + ……]
= (15/4) x^2 + (5/2) x^2 + ……
= (25/4) x^2 + ……

Hence, the coefficient of x^2 term in the expansion of (1 + x^2) [2x - (1/2x)]^6 = 25/4

I have assumed you meant 1 over 2x!
(1+x^2) (2x- 1/(2x))^6

Multiply (2x -1/(2x))^6 by 1
= (2x -1/(2x))^6
The coefficient of x^k is 6C(k-1) (2x)^(6-k+1) (-1/(2x))^(k-1)

= 6C(k-1) 2^(6-k+1) x^(6-k+1) (-1)^(k-1) 2^(-k+1) x^(-k+1)
We want the x^2 term:
x^(6-k+1-k+1) = x^2
6-2k+2 = 2
8-2k=2
-2k = -6
k=3

= 6C2 2^4 x^4 (-1)^2 2^(-2) x^(-2)
= (15) (16) (1) (1/4) x^2
= 60x^2
The coefficient of x^2 in 1*(2x -1/(2x))^6 is 60

Multiply (2x - 1/(2x))^6 by x^2
Because we need the coefficient of x^2 , what we need is the coeffcient of x^0 in (2x - 1/(2x))^6
= 6C(k-1) 2^(6-k+1) x^(6-k+1) (-1)^(k-1) 2^(-k+1) x^(-k+1)
We want the x^0 term:
x^(6-k+1-k+1) = x^0
6-2k+2 = 0
8-2k= 0
-2k = -8
k=4

= 6C3 2^(6-4+1) x^(6-4+1) (-1)^(4-1) 2^(-4+1) x^(-4+1)
= (20) 2^3 x^3 (-1)^3 2^(-3) x^(-3)
= -20 x^0

The coefficient of x^0 in (2x- 1/(2x))^6 = coefficient of x^2 in x^2 (2x -(1/(2x))^6 = -20

60 -20 = 40

The coefficient of x^2 in the expansion of (1+x^2)*(2x-1/2x)^6 is 40

(2x-1/2x)^6 is of the form (a+b)^6 which using Pascal's Triangle equals

1a^6 + 6a^5*b + 15a^4*b^2 + 20a^3*b^3 + 15a^2*b^4 + 6a*b^5 +1b^6. Here a=2x, b = -1/(2x).

The middle term 20a^3*b^3 is the term where the 2x and the -1/(2x) are going to cancel the x's, and equals 20(2x)^3 * (-1/(2x))^3 = -20.

So (1+x^2)*(2x-1/2x)^6 has -20 as the coefficient of x^2.

Check for a typo.
As written the answer is 0.
See: http://www.wolframalpha.com/input/?i=expand((1%2Bx%5E2)*(2x-1%2F2x)%5E6)