For the first part CRC will be 1001. So codeword will be 1100111001.
Now in CRC, If the generator is of degree d then it can detect all the burst error of length d, and all the burst error of length d+1 except one pattern.
so, in above bit sequence (1100111001) , burst error of length 5 can be generated like this:
0 _ _ _ 011001 .....(1)
(we can put either 0 or 1 in place of _ , as those bits may or may not be in error.)
OR 10_ _ _ 01001 .....(2)
OR 111_ _ _0001 ....(3) and more.
Now looking at the generator (11001), we can choose intermediate bits such that it becomes multiple of 11001.
so for Eq(1) we can choose M' as 0000011001
for Eq(2) 1010101001.
Please correct me if this is wrong.