It helps if you recognize the type of probability distribution that will assist you on your way. What you have here is Binomial with Probability of Success = 0.06 and a sample size of 6. I used 0.06, the % of finding a bad bolt on one draw, rather than 0.94, the probability of finding a good bold on one draw, because the questions are about bad bolts and it might be easier to understand. You should be able to use a table or a calculator simply to calculate the necessary values.
In my opinion, with a very small sample, one should simply construct the entire distribution and then answer ALL the questions. You can calculate the individual probabilities, but the construction of the entire distribution can be helpful and instructive. With a sample of 6, the only possible numbers of bad bolts are: 0, 1, 2, 3, 4, 5, and 6
The structure of the probability distribution is very pattern-intensive, so it's easy to construct. A spreadsheet can help with the tedious arithmetic.
p(0 bad bolts): 1 * 0.94^6 * 0.06^0 = 0.689869781056
p(1 bad bolts): 6 * 0.94^5 * 0.06^1 = 0.264205448064
p(2 bad bolts): 15 * 0.94^4 * 0.06^2 = 0.04216044384
p(3 bad bolts): 20 * 0.94^3 * 0.06^3 = 0.00358812288
p(4 bad bolts): 15 * 0.94^2 * 0.06^4 = 0.00017177184
p(5 bad bolts): 6 * 0.94^1 * 0.06^5 = 0.000004385664
p(6 bad bolts): 1 * 0.94^0 * 0.06^6 = 0.000000046656
The only slightly tricky part is those magic numbers out front. They are called "Binomial Coefficients". They are not very difficult to calculate. Notice that they appear in a palindrome pattern. This helps, because you must calculate only about half of them.
Once we have the whole distribution, the questions are simple:
(i) Two of the six bolts exceed the diameter.
p(2 bad bolts): 15 * 0.94^4 * 0.06^2 = 0.04216044384
(ii) More than two of the six bolts exceed the diameter.
p(3 bad bolts): 20 * 0.94^3 * 0.06^3 = 0.00358812288
p(4 bad bolts): 15 * 0.94^2 * 0.06^4 = 0.00017177184
p(5 bad bolts): 6 * 0.94^1 * 0.06^5 = 0.000004385664
p(6 bad bolts): 1 * 0.94^0 * 0.06^6 = 0.000000046656
