The Language L= {aⁿ bⁿ | n>=0} is CFL but not regular, as it requires comparison between a’s and b’s.
L = {aⁿ | n is prime} is CSL, as calculation of “n is prime” can be done by LBA (Turing machine)
L = {ww | w ∈ ∑*} is CSL.
But L = { w | w has 3k+1 b’s for some k ∈ natural number} is regular.
Lets take values of k={1,2,3,….}
So number of b’s will be {4, 7, 10,……….} and number of a’s can be anything.
The DFA will be

The regular expression equivalent to L1 and L2 are (a*) and (b*) respectively.
Since L1 and L2 both are regular languages and regular languages are closed under complementation, so there concatenation (i.e. L1. L2) must also be a regular language.
But,
L1.L2 ≠ { aⁿbⁿ | n ≥ 0}
L1= {ϵ, a ,aa, aaa, aaaa,.............}
L2={ϵ, b, bb, bbb, bbbb, …………}
So L1.L2 = {ϵ, a ,aa, aaa, aaaa,.............}.{ϵ, b, bb, bbb, bbbb, …………}
L1.L2= {ϵ, a, aa, aaa, …………., b, bb, bbb, ……., ab, abb, abbb, …….., aab, aaab, …………..}
Hence L1.L2= { a^mbⁿ | m, n ≥ 0}

In L₁ any string “w” must satisfy the condition:
{Number of occurrences of (110)} ≥ {Number of occurrences of (011)}
Lets analyse the language, consider a string in which occurrence of (110) is more than one.
The following possibilities are: {1100110, 1101110, 110110, ….}
Please observe whenever strings start with “11” then in every situation whatever comes after “11” the string will never violate the condition. So strings of the form 11(0+1)* will always satisfy the condition.
Consider a string in which occurrence of (011) is more than one.
The following possibilities are: {011011, 0111011, 0110011, ….}
In the following possibilities please observe that number of occurrence “011” is two but number of occurrence of (110) is one, which violate the conditions.
If we add “0” in every string mentioned above, i.e. {0110110, 01110110, 01100110, ….} Now, observe that number of occurrence “011” and the number of occurrence of (110) both are equal, which satisfies the conditions.
With these analysis, we can make the DFA , which is mentioned below.

But language L₂ requires infinite comparison to count the occurrences of (000’s) and (111’s), hence it is not regular.

Statement I represents a regular language whose regular expression is a* (bb)*. Also it doesn’t require any comparison between “a” and “b” , so it can be recognized by DFA and hence regular.
Statement II and III represent CFL, as it requires comparison between number of a’s and b’s.
Statement IV is also regular, and its regular expression is (a+b)* c (a+b)*.

Regular languages are not closed under subset.

(I) is the correct definition and the other two is wrong because the other two can have any no. of x and y. There is no such restriction over the number of both being equal.

L∪R is nothing but L itself. Because R is subset of L and hence regular. R is deterministic context free but not regular as we require a stack to keep the count of 0's to make that of 1's.