I am stuck on "part (a)" of this question.
My approach:
X, Y and Z are independent Bernoulli random variables.
So, XYZ is also a Bernoulli random variable, since it can take only two values - {0,1} (1 when all of them are 1 and 0 when any of X, Y or Z is 0)
Therefore P(XYZ = 0) = 1 - {0.5*0.5*0.5} = 7/8
Now, 7/8 = 0.875, which is greater than 3/4 or 0.75.
But I don't think this is the right approach to solving this question, since I wasn't able to arrive at the inequality explicitly.
I would be very grateful, if somebody can help me with this question.
Oh, no, you were informed that the variables are pairwise independent, but that does not assure you that they will be mutually independent.
So you cannot justify saying: $\mathsf P(XYZ{=}1)=\mathsf P(X{=}1)\mathsf P(Y{=}1)\mathsf P(Z{=}1)$. It may not be true (and indeed, part (b) indicates that it will not be true).
However, you can say $\mathsf P(XYZ{=}0) = \mathsf P(X{=}0\cup Y{=}0\cup Z{=}0)$ and then use the principle of inclusion and exclusion to express this as a series of seven terms. The information given will allow you to evaluate six of these seven terms, and there is an inequality you do know about the seventh propability measure.
Go to it.