Instead of testing all divisors in range(2,num), you could extract the test for even numbers and then loop only on the odd numbers. Additionally, as Bill the Lizard suggests, you could stop at the square root of num. This will be twice as fast:
def eprimo(num):
if num < 2:
return False
if num % 2 == 0:
return num == 2
div = 3
while div * div <= num:
if num % div == 0:
return False
div += 2
return True
