Answer
a)0.0043
b)0.00036
c)0.00479
d)Yes, it is unusually low.
Work Step by Step
a) $P(X=2)={7\choose 2} \cdot (0.2)^5\cdot (0.8)^2=0.0043.$
b)$P(X=1)={7\choose 1} \cdot (0.2)^6\cdot (0.8)^1=0.00036.$
c)$P(X<3)=P(X=0)+P(X=1)+P(X=1)={7\choose 0} \cdot (0.2)^7\cdot (0.8)^0+{7\choose 1} \cdot (0.2)^6\cdot (0.8)^1+{7\choose 2} \cdot (0.2)^5\cdot (0.8)^2=0.00013+0.00036+0.0043=0.00479$
d) Yes, it is an unusually low probability, because 0.00479 is really low, lower than 0.05.