Both of the them are regular
Here S2 = {anbn | n>=1} ∪ { anbm | n>=1,m>=1 } = { anbm | n>=1,m>=1 } which is regular.
And S1 = { (an)m | n <= m >= 0 }
Let an = b then S1 becomes
S1 = { bm | n <= m >= 0 } which is also regular.
Look at the image below
Here FIG A gives the DFA for S2 and FIG B gives the DFA for S1.