I know these problems were posted before, but I found a more intuitive approach to solving them that I think might be helpful...

9747: Let x and y be uniformly distributed random variables on [0,1] the probability that the distance between x and y is less than 1/2 is...

To solve this we note that the restriction requires |x-y| < 1/2, so that x - y < 1/2 or x-y > -1/2. These give y>x-1/2 and y<x+1/2. The solution is to graph the lines y=x-1/2 and y=x+1/2 in the square region [1,0]x[0,1] and calculate the area between them. The probability that |x-y| < 1/2 is equal to this area, which is 3/4.

9340: If x,y and z are selected at random from the interval [0,1], then the probability that x >= yz is...

In this case, we graph the region x=yz in the cube [1,0,0]x[0,1,0]x[0,0,1] and then compute the area in the region for which x > yz. This is the area above the curve x=yz, which is the volume of the cube minus the volume integral of yz with all bounds going from 0 to 1. The probability that x >= yz is therefore equal to 1 - 1/4 = 3/4.