Python 3.5, 280 272 260 242 240 bytes:
(Thanks to Adnan for the trick about using the * operator in comparisons resulting in 2 saved bytes!)
def g(o):import urllib.request as u,re;R=re.findall;w=bytes.decode(u.urlopen('http://ppcg.lol/q/'+o).read());print((len(R('(?:<h[0-9]>|<p>).*python',w.lower()))<2)*(int(R('(?<="vote-count-post ">)[0-9]+',w)[0])>3)*w.count('answercell">')>5)
Simple enough. Uses Python's built in urllib library to go to the question's site, and then uses regular expressions to find the vote count, answer count, and the count of Python specific answers in the decoded text returned from the website. Finally, these values are compared against the conditions required to return a truthy value, and if they satisfy all the conditions, then True is returned. Otherwise False is.
The only thing I may be worried about here is that the regular expressions give a lot of lee way in the terms of the number of python specific answers to save bytes, so it may be a bit inaccurate at times, though it's probably good enough for the purposes of this challenge. However, if you want a much more accurate one, I have added one below, although its longer than the one above. The one shown below is currently 298 bytes since it uses a much longer regular expression–one which you couldn't know how long it took me to discover–for counting Python answers than my original function for the sake of accuracy. This one should work for about at least 80% to 90% of all test cases thrown at it.
def g(o):import urllib.request as u,re;R=re.findall;w=bytes.decode(u.urlopen('http://ppcg.lol/q/'+o).read());print(len(R('(?<=answercell">).*?(?:<h[0-9]>|<strong>)[^\n]*python[^\n]*(?=</h[0-9]>|</strong>)',w.lower()))<2and int(R('(?<="vote-count-post ">)[0-9]+',w)[0])>3and w.count('answercell">')>5)
But, what about those questions with multiple pages of answers? Neither of the above will work very well in that situation, if, say, 1 python answer is on the first page and another is on the second. Well, I took the liberty to fix this issue by creating another version of my function (shown below) that checks every page of answers, if multiple ones exist, for Python answers, and it has done quite well on many of the test cases I have thrown at it. Well, without further ado, here is the new and updated function:
def g(o):
import urllib.request as u,re;R=re.findall;w=bytes.decode(u.urlopen('http://ppcg.lol/q/'+o).read());t=0if len(re.findall('="go to page ([0-9]+)">',w))<1else max([int(i)for i in re.findall('="go to page ([0-9]+)">',w)])
if t<1:print(len(R('(?<=answercell">).*?(?:<h[0-9]>|<strong>)[^\n]*python[^\n]*(?=</h[0-9]>|</strong>)',w.lower(),re.DOTALL))<2and int(R('(?<="vote-count-post ">)[0-9]+',w)[0])>3and w.count('answercell">')>5)
else:
P=[];U=[];K=[]
for i in range(2,t+2):P.append(len(R('(?<=answercell">).*?(?:<h[0-9]>|<strong>)[^\n]*python[^\n]*(?=</h[0-9]>|</strong>)',w.lower(),re.DOTALL)));U.append(int(R('(?<="vote-count-post ">)[0-9]+',w)[0]));K.append(w.count('answercell">'));w=bytes.decode(u.urlopen('http://ppcg.lol/questions/'+o+'/?page='+str(i)).read())
print(sum(P)<2and U[0]>3and sum(K)>5);print('# Python answers: ',sum(P));print('# Votes: ',U[0]);print('# Answers: ',sum(K))
Quite long, isn't it? I wasn't really going much for code golf with this, although, if you want, I can golf it down a bit more. Otherwise, I love it, and could not be happier. Oh, I almost forgot, as an added bonus, this also outputs the total number of Python answers on the question, total votes on the question, and total number of answers on the question if the question id corresponds to a question with more than 1 page of answers. Otherwise, if the question only consists of a single page of answers, it just outputs the truthy/falsy value. I really did get a bit carried away with this challenge.
These each take the question's id in the form of a string.
I would put Try It Online! links here for each function, but unfortunately, neither repl.it nor Ideone allow fetching of resources via Python's urllib library.
