Let $R_1,R_2$ and $R_3$ be the three rooms and let $A, B ,C$ denote the three friends.
I represent the white balls by the digit $1$ and the black ones by $0$.
They could adopt the strategy as summarized in the table below where I have indicated all the possibilities where the $plus$ and $dot$ symbol represent the Boolean operators OR and AND respectively.The overhead operator indicates NOT operation.
In this strategy $C$ always guesses $1$ (White ball), $A$ and $B$ guess based on what $R_3$(room 3) has. Their guesses are determined by simple Boolean algebra.
So for eg in the first case i.e in a) first A looks in $R_3$ and on seeing a white ball he performs the operation $R_2 .R_3$ meaning $1.1 = 1$(as per Boolean algebra). $A$ guesses correctly. Next $B$ will perform the NAND operation of $R_1$ and $R_3$ which gives zero.Here $B$ makes an incorrect guess and finally C will simply say white every-time(correct for this case). So they get 2 out of 3 correct. Rest of the table is self-explanatory.
The red ones indicate the cases which would fail. So they can win 6 out of 8 times.
Also this method works for the case when all balls have the same colour.
For Variation 1
Here the strategy should be to guess all those combinations correctly which has more combined probability of occurrence. The table below shows one such strategy where $C$ always guess $0$ (Black ball). $A$ guess $1$ if $R_2 = 1$ otherwise $0$. $B$ guess $1$ if $R_1 = R_3$ otherwise $0$.
This way they manage to guess the 6 out of 8 which has a total probability of 0.874.
Ideally we would want them to not include the combination $(1,1,1)$ which has the minimum combined probability and you would want to guess $(0,0,0)$ combination for sure. But I don't think that's possible. Either you can have both $(1,1,1)$ and $(0,0,0)$ or neither of them in your correct 6. I'm not sure of this,maybe there is some other way.
Edit : After reading the comments I realized that i was wrong about $(1,1,1)$ and $(0,0,0)$ combination.
But I think that this is the optimal solution which gives the maximum total probability of occurrence for 6/8 because there is only one other combination which gives a higher value (0.91) i.e. when you exclude the $(1,1,1)$ + any other with two $1's$ viz ((a,b) , (a,c) or (a,e)). But if you choose to exclude any of these two then you would surely fail.
Like for eg. if you try to exclude (a,b) then you need to satisfy the following:-
Firstly the two failed one's should necessarily have all guesses incorrect. With this constraint we run into contradiction when two same initial configuration demand exactly opposite result(see blue coloured in table). Similar contradiction happen for the other two cases. So by contradiction we cant have the 0.91 solution so the next highest has to be the optimal solution.