Answer:
a.
CFG for {

, m=n-1 and n=1,2,3… …}
Here we can have a string containing only a single a and no b.
For that the production rule is S → aT, T → ε
Now to get the strings with at least one b, the production rule is to be
T → ATB/ ε
A → a
B → b
Now merging two set of production rules:
S → aT
T → ATB/ε [As T → ε is in both set, so one occurrence is taken]
A → a
B → b
Now let us generate “aaabb”
S → aT → a ATB → aAATBB → aaATBB → aaaTBB → aaaεBB → aaaεbB →aaaεbb → aaabb
b.
CFG for {

, n=1,2,3… …}
Here for every single a, we have to generate two b’s.
So the production rule is to be:
S → aSbb/ε
Each time S will generate two b’s for one a.