The easiest way I know to answer your question about the dice is to use a normal approximation! The answer is that the probability that Julie’s sum is greater is about 0.745.
First let’s consider the result of rolling one 8-sided die. Let’s say the result is U. U has a “discrete uniform distribution” (see Uniform distribution (discrete) - Wikipedia, the free encyclopedia) so its mean is (1+8)/2 = 4.5 and its variance is (8^2 – 1) / 12 = 5.25.
Julie’s roll, say J, is the sum of 4 of the U random variables, so the mean of J is 4 * 4.5 = 18 and its variance is 4 * 5.25 = 21.
Hasish’s roll, say H, is the sum of 3 of the U random variables, so the mean of H is 3 * 4.5 = 13.5 and its variance is 3 * 5.25 = 15.75.
Define Y = J-H. We are interested in Pr(Y > 0). The exact distribution of Y is hard to work with, so let’s try a normal approximation. This seems reasonable because Y is the sum of several independent random variables (the U’s or their negatives). We know the mean of Y is 18 – 13.5 = 4.5 and its variance is 21 + 15.75 = 36.75, so its standard deviation is sqrt(36.75) = 6.06. Define Z = (Y – 4.5) / 6.06; then Z has mean 0 and standard deviation 1, so if we assume Z has a normal distribution (it doesn’t really, but it’s close) then we can look up the cumulative distribution in a table. Here we use a little trick to improve the accuracy of the approximation: we want to know Pr(Y>0), which is the same as Pr(Z > -4.5/6.06). But the distribution of Y is discrete and the distribution of Z is continuous, so let’s consider Pr(Y > 0.5) instead. This trick is so common that it has a name, the “correction for continuity”. Pr(Y > 0.5) = Pr(Z > -4.0/6.06 = -0.660), and looking up the cumulative normal distribution in a table we find this probability is 0.745. That’s the answer to your problem, approximately.
You might ask how accurate is the approximation. I used more advanced methods involving generating functions to find the exact distribution of Y and found that Pr(Y>0) = 1,554,908 / 8^7, which is about 0.7414. This would have been very tedious and time-consuming to find except that I used a computer algebra system to do the hard work. So in this case the normal approximation is pretty good, with a relative error of only about 0.5%.