All are TRUE.
I. If A is regular, then B is regular: Given A is regular, that means A can be written as a regular expression. Say, it's regular expression is e. Then the regular expression for B will be: (a+b)*e(a+b)*. Since we have a regular expression for B, this implies that B is also regular.
II. If A is recursive, then B is recursive: Given A is recursive this means that there exists a TM which given a string as input can tell whether it belongs to A or not. Lets call it TM_A. Now, using this we can create a TM which can be used for telling if a string belongs to B or not(calling it TM_B). This will prove that B is also recursive.
How to create such a TM_B?
When we have an input string for TM_B, take every possible substring of it and given it to TM_A, if any of the substring is accepted by TM_A, then the original string belongs to B. Otherwise if none of the substring is accepted by TM_A, that means that the original input string does not belong to B. Every string will be of finite length, that means that there will be a finite number of substrings to check. So, we have a TM for B, which always tells if a string is in B or not. Hence, B is also recursive.
III. If A is context-free, then B is context-free: A is context free then there will be a CFG representing it. Let's say its starting symbol is S. Now we can create another CFG with starting symbol S'
S' -> ASA
A -> aA | bA | e
Here S is the starting symbol of A's CFG. This will produce all the strings that have some string of A as substring.
There is a CFG for B, therefore B is context free.
ALL ARE TRUE!!!