Turing Machine
Question 1 
L_{1} is recursive and L_{2}, L_{3} are not recursive  
L_{2} is recursive and L_{1}, L_{3} are not recursive  
L_{1}, L_{2} are recursive and L_{3} is not recursive  
L_{1}, L_{2}, L_{3} are recursive 
Question 1 Explanation:
L_{1} is recursive: Since counting any number of steps can be always decided. We can simulate TM (M) whether it takes more than 2016 steps on some input string, which has length upto 2016. If it happens then reached to accepting (YES) state else reject (NO).
L_{2} is recursive: Similarly, we can simulate TM (M) whether it takes more than 2016 steps on each input string which has length upto 2016, If it happens then reached to accepting (YES) state else reject (NO).
L_{3} is not recursive: If L_{3} is recursive then we must have a Turing machine for L_{3}, which accept epsilon and reject all strings and always HALT. Since Halting of Turing machine can’t be guaranteed in all the case, hence this language is not recursive.
L_{2} is recursive: Similarly, we can simulate TM (M) whether it takes more than 2016 steps on each input string which has length upto 2016, If it happens then reached to accepting (YES) state else reject (NO).
L_{3} is not recursive: If L_{3} is recursive then we must have a Turing machine for L_{3}, which accept epsilon and reject all strings and always HALT. Since Halting of Turing machine can’t be guaranteed in all the case, hence this language is not recursive.
Question 2 
Let <M> be the encoding of a Turing machine as a string over Σ = {0, 1} Let L = {<M> M is a Turing machine that accepts a string of length 2014}. Then, L is
decidable and recursively enumerable  
undecidable but recursively enumerable  
undecidable and not recursively enumerable  
decidable but not recursively enumerable 
Question 2 Explanation:
If L is recursive language then there must exist a Turing Machine which always HALT for every case (either acceptance or rejectance of string). Let the Turing Machine for L is M1.
Now, since total number of strings of length 2014 is finite, so M1 will run the encoding of M for the string of length 2014 and if the M accept the string then M1 will halt in ACCEPT state. But if M goes for infinte loop for every string of length 2014, then M1 also will go into infinite loop. Hence language L is recursively enumerable but not recursive, as in case of rejectance halting is not guaranteed.
Now, since total number of strings of length 2014 is finite, so M1 will run the encoding of M for the string of length 2014 and if the M accept the string then M1 will halt in ACCEPT state. But if M goes for infinte loop for every string of length 2014, then M1 also will go into infinite loop. Hence language L is recursively enumerable but not recursive, as in case of rejectance halting is not guaranteed.
Question 3 
M does not halt on any string in (0+1)^{+}
 
M does not halt on any string in (00+1)*  
M halts on all strings ending in a 0
 
M halts on all strings ending in a 1

Question 3 Explanation:
Try for any string, it will not Halt for any string other than ϵ. Hence, option (A) is correct.
There are 6 questions to complete.