To find the intervals where the expression is greater than 0, we need to determine the sign of the expression in each interval and use the properties of inequality.

First, let's find the critical points by setting each factor equal to zero and solving for x:

(2x+1) = 0
2x = -1
x = -1/2

(1-2x) = 0
-2x = -1
x = 1/2

(x-1) = 0
x = 1

(2-3x) = 0
-3x = -2
x = 2/3

Now, let's test each interval:

Interval 1: (-∞, -1/2)
Pick x = -1:
(-)(+)(-)(+) > 0 is true

Interval 2: (-1/2, 1/2)
Pick x = 0:
(+)(+)(-)(+) > 0 is false

Interval 3: (1/2, 1)
Pick x = 1:
(+)(-)(+)(+) > 0 is true

Interval 4: (1, 2/3)
Pick x = 3/4:
(+)(-)(+)(-) > 0 is false

Interval 5: (2/3, ∞)
Pick x = 1:
(+)(-)(+)(-) > 0 is true

Therefore, the solution to the inequality (2x+1)(1-2x)(x-1)(2-3x) > 0 is x ∈ (-∞, -1/2) U (1/2, 1) U (2/3, ∞).