Answer
a) 120
b) 24
c) 120
d) 24
e) 6
Work Step by Step
a) We treat the string BCD as single letter, thus we have then the possible letters A, BCD, E, F, G. So, The problem is to count permutations of five items ~ the letters A, E, F, G and BCD. Therefore, the answer is
$$^5P_5= 5! = 120.$$
b) We treat the string CFGA as single letter, thus we have then the possible letters B, D, E, CFGA. So, The problem is to count permutations of four items ~ the letters B, D, E and CFGA. Therefore, the answer is
$$^4P_4= 4! = 24.$$
c) We treat the string BA as single letter and GF as another string, thus we have then the possible letters BA, C, D, E, GF. So, The problem is to count permutations of five items ~ the letters BA, C, D, E and GF. Therefore, the answer is
$$^5P_5= 5! = 120.$$
d) We treat the string ABC as single letter and DE as another string, thus we have then the possible letters ABC, DE, F, G. So, The problem is to count permutations of four items ~ the letters ABC, DE, F and G. Therefore, the answer is
$$^4P_4= 4! = 24.$$
e) If both ABC and CDE are substrings, then ABCDE has to be a substring. So we are really just permuting three items: ABCDE, F, and G. Therefore the answer is
$$^3P_3= 3! = 6.$$
f) There are no permutations with both of these substrings, since B cannot be followed by both A and E at the same time.
