Find primes that are also primes when its digits are rotated
A 2+ digit fare prime number can never have an even digit, or a 5 digit in its set of digits, because a rotation which moves an even digit or a 5 to the last digit will be divisible by 2 or 5.  You could use that as a filter for your possible fare primes from the list of primes you calculate. 

When calculating primes, you can stop at sqrt(num).  Any number greater than sqrt(num) that evenly divides num will have a compliment number less than sqrt(num) that you have already tested. 

Speaking of primes that you’ve calculated, why don’t you use those for your subsequent prime test?  Why try dividing by every number from 2 up to int_num when you could just try the numbers in primes upto int_num.

... or just ask if int_num in primes.  Speed tip: turn primes into a set() first for faster inclusion testing.

Your digit_list code is very inefficient. For any number, once you’ve split the number into digits, you don’t need to resplit it into digits again for each rotation.  Actually, you don’t even need to split it into individual digits.  This will give you the rotated values:

digits = str(i)
for j in range(1, length):
    int_num = int( digits[j:] + digits[:j])