Solution
From (A):
$ \text{Some green are blue} \Rightarrow \text{intersection of green and blue is non–empty} $
So we can also say:
$ \text{Some blue are green} $
Therefore Conclusion I follows.
From (B):
$ \text{No blue is white} \Rightarrow \text{all blue are non–white} $
Combine with (A):
Those objects which are both green and blue are also not white.
So:
$ \text{Some green are not white} $
Therefore Conclusion III also follows.
For II (Some white are green) and IV (All white are green),
nothing in the statements links white to green positively. Both may or may not overlap, so these do not follow.
Hence only I and III are valid.