for part a) entropy will decrease because if V2is smaller than V1, inside the brackets is a fraction, this will make ln(answer) a negative number and therefore the whole right side negative.
for part b) entropy will increase because if V2is larger than V1, inside the brackets is larger than 1, this will make ln(answer) a positive number and therefore the whole right side positive.
Another way to think about it logically is to think of the definition of entropy- measure of disorder or the number of ways or arranging the molecules of gas (in this case). If you compress the gas (part a) then there are less ways to arrange the molecules. If you expand, more ways to arrange.
for part c) make V the subject of the ideal gas equation: V= (nRT)/p. And since n,R and T are the same for both pressure/volume you can sub this into the change in entropy equation to get: ΔS = nR ln[(nRT)/p2/] / [(nRT)/p1] sorry for how that looks btw. The nRT cancel out leaving you with (1/p2) / (1/p1) flip (1/p1) around to make : (1/p2)*p1 which equals (p1/p2). Therefore the full equation:
ΔS = nR ln(p1/p2).