Most of the time, GMAT Geometry problems can be solved by more than one ways. That is the beauty of geometry.

I go like this:

Problem 1:
We have following information; radius (9), angle ORP = 35, Lines PQ and OR are parallel. Let C be the center of the circle. So basically we need to find the ∠QCP to find the length of minor arc PQ.

Since lines PQ and OR are parallel, ∠ORP and ∠QPR are same, 35 each.
Tringle PCR is an isosceles triangle and so base angles are same. That means ∠CRP and ∠CPR are 35 each. Similarly for the Triangle QCP is an isosceles triangle and so ∠CPQ is 70 (35+35) and so does ∠CQP. That means ∠QCP is 40.

So length of minor arc PQ (say, x) can be calculated as follows.

x= 40*2*9*pi/360 = 2*pi

Answer is: A

Problem 2:

The length OP or the radius of the semicircle is = SQRT{(sqrt3)^2+1^2} = SQRT{4} = 2

Now consider a triangle by drawing a straight line between P and Q. Since the ∠POQ is 90 and two sides of the triangle OPQ are equal, the distance between O and Q is same as OP = 2, and its base angles are 45 each. Using the 45:45:90 ratio of angle, the length PQ is basically 2*sqrt(2), which is the sum of x-axis distance of P from O [i.e. sqrt(3)] and x-axis distance {s} of Q from O.

In other words, 2*sqrt(2) = sqrt(3) + s.

The distance s (x value) = 2*sqrt(2) - sqrt(3) = 2*1.41 - 1.73 = 1.08

Hence the answer is: B

I am attaching a pdf file with explanation.
Thank You!