What is the updated belief of B given F and I with all of them being independent?
This is more of a smell check cause I'm decently sure I have the correct thought process but it seems weird in the context of the story:
"Say you are 20% sure GameStop stocks will go up in a week, but your friend says it will for sure go up and your friend is correct 60% of the time. Then say you talk to a professional investor and they say that the stock will go up and is correct 80% of the time. Assume your belief, your friend's accuracy and the investor's accuracy are all independent events. What is your updated belief after observing what your friend says? How about after observing your friend and the investor?"
So I'm decently sure it's 20% for both cases, because the events are independent. So let's call my belief event B, friend's correctness F and investor's correctness as I. Then $P(B|F)=\frac{P(B,F)}{P(F)}$, but since they're independent $P(B|F)=\frac{P(B)P(F)}{P(F)}=\frac{(0.2)(0.6)}{0.6}=0.2$ then also because I is independent of B and F then $P(B|F,I)=\frac{P(B,F,I)}{P(F,I)}=\frac{P(B)P(F)P(I)}{P(F)P(I)}=P(B)=0.2$. This all makes sense to me given that they're all independent and only the marginal probabilities are given, it'd be impossible to solve this if they weren't independent because the joints and conditionals are not given, but it feels weird that my belief wouldn't improve given my friends and investor's correctness as evidence.
Call $S$ the unknown event (i.e. that the stock is actually going to go up) and $A,B,C$ the events for the people's "hunches" (including your own, if you want).
It's reasonable to treat $A,B,C$ as independent given $S$, i.e. $P(A,B,C|S)=P(A|S)P(B|S)P(C|S)$, and you can model the different people's success rates using the conditional probabilities for each of the hunches (so e.g. $P(C|S)=P(\lnot C|\lnot S)=0.8$). Doing Bayesian inference to take the hunches into account means calculating $P(S|A,B,C)$, and this can definitely be different from $P(S)$ (and from $P(S|A)$, where $A$ is your own hunch).
With the information you gave, I'd leave out $A$ (since $P(S)$ already represents your prior belief in $S$). Then: $$ \begin{align} P(S|B,C) &=\frac{P(S,B,C)}{P(B,C)}\\ &=\frac{P(S,B,C)}{P(S,B,C)+P(\lnot S,B,C)}\\ &=\frac{P(S)P(B|S)P(C|S)}{P(S)P(B|S)P(C|S)+P(\lnot S)P(B|\lnot S)P(C|\lnot S)}\\ &=\frac{(.2)(.6)(.8)}{(.2)(.6)(.8)+(.8)(.4)(.2)}\\ &=0.6 \end{align}$$
So hearing your friends' predictions increases your confidence from $0.2$ to $0.6$.