I)Set of all strings containing any number of ‘a’ s followed by an even number of ‘b’ s. R.E=(a)$^{*}$(bb)$^{*}$.
IV) Strings containing a ‘c’. R.E= (a+b)$^{*}$c(a+b)$^{*}$.
Both these languages are regular as regular expressions exist.
By default a language is infinite. Eg : {a$^{n}$} it’s a infinite language.So both the languages II and III are infinite and comparison has to be done to evaluate these and hence are not regular.
Answer: A
NOTE:
Every finite language is regular.
Infinite language + Comparison = Non-Regular.
Infinite language + No Comparison = Regular.
Edit: As nothing is mentioned about ‘c’ in option IV and there is a comma after y, So I think It’s a typo ‘c’ should also belongs to {a,b}$^{*}$. IV will be a complete language. Which is regular. R.E=(a+b)$^{*}$.