That's the set of all elements of B that are not elements of B

From B = {4,5,6,7}, we take out any of these A={1,2,3,4,5} which
B contains.  A contains 4 and 5, which A also contains, so we 
remove 4 and 5 from B and we have {6,7}.

Therefore B-A = {6,7}


                                                          
(ii)(A-D)c

We first find what's in the parentheses, A-D

From A={1,2,3,4,5}, we take out any of these D={2,4,6,8} which
A contains.  D contains 2 and 4, which A also contains, so we 
remove 2 and 4 from A and we have {1,3,5}.

So A-D = {1,3,5}

So to find the complement of that set, (A-D)c,
we form the set of all elements in the universal set U={1,2,3,…,9}
which are not elements of A-D, which is {1,3,5}. So we remove
1,3 and 5 from the universal set U={1,2,3,…,9} and we have
{2,4,6,7,8,9}

So (A-D)c = {2,4,6,7,8,9}

We are only supposed to answer 2 questions per post.

Edwin