Here's my version. I decided to store the primes I'd already calculated in a list and use them as the divisors for the next candidate, rather than determining the divisors each time through the loop. Python's high-level, I imagine its list implementation is efficient enough for me not to worry about it :)
A number can't have a factor that's greater than its square root, so if it has reached that point, it's a prime and the control can break out of the loop.

#Problem 1
primes = [2] #list of primes. We know that 2 is the first prime
integer = 3 #so start checking at 3

while len(primes) < 1000:
   for p in primes: #we want to divide the candidate integer by all the primes we already have
       is_prime = 1 #we'll assume it's prime until proven otherwise
           if (integer % p == 0):
	       is_prime = 0 #if it can be divided equally by a prime, it's not prime
	       break        #so break out of the loop here
	   if p*p >= integer:
		break #don't keep checking with divisors greater than the square root
    if is_prime: #stopped dividing; is it prime?
        primes.append(integer) #add it to the array of primes
    integer = integer + 1 #and then move on to the next integer
#we're out of the while loop, so print the last prime, 
#and let's check how many we have while we're at it
print "The %dth prime is: %d" %(len(primes), primes[len(primes)-1]) #Don't forget the list is 
#zero-indexed, or we'll get a lovely index out of bounds error here

And the slightly adapted version for the second problem:

from math import *
n = int(raw_input("Enter a number: ")) #ask the user what is n
primes = [2] #list of primes. We know that 2 is the first prime
integer = 3 #so start checking at 3
sum_logs = log(2) #the log of the primes we have so far

while len(primes) < n:
    for p in primes: #we want to divide the candidate integer by all the primes we already have
        is_prime = 1 #we'll assume it's prime until proven otherwise
            if (integer % p == 0):
        is_prime = 0 #if it can be divided equally by a prime, it's not prime
        break        #so break out of the loop here
        if p*p >= integer:
            break #don't keep checking with divisors greater than the square root
     if is_prime: #stopped dividing; is it prime?
        primes.append(integer) #add it to the array of primes
        sum_logs = sum_logs + log(integer) #find the log and add it to the total
    integer = integer + 1 #and then move on to the next integer
#print out the sum of the logs, and the ratio
print "The sum of the logs of the first",len(primes), "primes is",sum_logs
print "The ratio of the sum to",n,"equals",(sum_logs / n)

I don't think I've put so many comments in code since I was in first year of uni ;)
You might also notice that I like to put double quotes around strings, and have a space on each side of the operators. I just think it's easier to read that way.