A die with 6 sides is rolled and the number on top is observed. What is the probability of rolling a die once and observing a number less than 5?1/62/35/64

so P(a 6 in two rolls) = P(6 in first roll) + P(6 in second roll) - P(6 in both rolls)
           = 1/6+1/6-1/6*1/6
           = 11/36

You could have also used:
P(a 6 in two rolls)  = P(6 in the first roll only) + P(6 in the second roll only) + P(6 in both rolls) 
                        = 1/6*5/6 + 5/6*1/6 + 1/6*1/6
                        = 11/36
   
The above approach is not easy to handle when the number of rolls are increased.

for two rolls:
for three rolls:
and so on.. Inclusion–exclusion principle 

Fortunately there is a better approach in this case, which will be clear if you see the venn diagrams in the link above, 
P(a 6 in two rolls) = 1 - P(no 6 in two rolls) 
P(no 6 in two rolls)= P(no 6 in first roll) * P(no 6 in second roll) 
and P(no 6 in first roll) = P(no 6 in second roll) = 5/6
from there you get:
P(a 6 in two rolls) = 1-(5/6)^2 
similarly for six rolls it will be 1-(5/6)^6 so its 'c'