We can use Rice theorem Part II to answer those.
Rice 2nd Theorem:- Any non monotonic property of the language recognizable by turing machine is unrecognizable.
A property P is said to be monotonic property of Recursively enumerable language if and only if P is satisfied by the Recursively enumerable language (L) then P should be satisfied by every Recursively enumerable language which is superset of L.
Option 1:-
$L$={<M> | M is a TM and |L(M)| $\leq$ 2022}
here the Property is “has at most 2022 string”
$ \sum = \left \{ a,b \right \}$
$TM_{yes}$ = $\left \{ \phi \right \}$
$TM_{No}$=$ \sum ^{*} $
now property P is satisfied for $TM_{yes}$ but it is not satisfied for $TM_{No}$ where $TM_{No}$ is superset of $TM_{yes}$ . So,It is non monotonic property of the language . So it is non recognizable by rice’s theorem.
Option 2:-
$L$={<M> | M is a TM and |L(M)| $= 2022$}
here the Property is “has exactly 2022 string”.
$ \sum = \left \{ a,b \right \}$
$TM_{yes}$=$\left \{ a_{1},a_{2},a_{3},.......,a_{2022} \right \}$ [$a_{i}$ represent i a’s concatenated where i=1 to 2022 ,eg. $a_{3}$=”aaa” ]
$TM_{No}$=$ \sum ^{*} $
now property P is satisfied for $TM_{yes}$ but it is not satisfied for $TM_{No}$ where $TM_{No}$ is superset of $TM_{yes}$ . So,It is non monotonic property of the language . So it is non recognizable by rice’s theorem.
Option D:-
$L$={<M> | M is a TM and L(M) $=$ {0,1}}
here the Property is “has only two strings “$0$,”$1$”.
$ \sum = \left \{ 0,1 \right \}$
$TM_{yes}$=$\left \{ 0,1 \right \}$
$TM_{No}$=$ \sum ^{*} $
now property P is satisfied for $TM_{yes}$ but it is not satisfied for $TM_{No}$ where $TM_{No}$ is superset of $TM_{yes}$ . So,It is non monotonic property of the language . So it is non recognizable by rice’s theorem.
Option B:-
$L$={<M> | M is a TM and |L(M)| $\geq$ 2022}
here the Property is “has at least 2022 string”
Now ,
If you think carefully you will find that if there is a $TM_{yes}$ has the property has at least 2022 string then any superset of $TM_{yes}$ also have at least 2022 string and satisfy the property. So it is monotonic property .
Now Can we say that as non monotonic property of the language recognized by TM is unrecognizable then monotonic property of the language recognized by TM is recognizable?
=> Definitely not.
Here The property is monotonic so we can’t apply Rice theorem here as it is only applicable for non monotonic property but not for monotonic property .
To proof that this is turing recognizable we use one technique named dovetailing.Here first should enumerate all the strings and give input to TM . First 0-length string,then 1-length string,2-length string and so on fed the turing machine machine in an interleaved manner like we will run one string for some time another string in a round robin way/interleaved way. Now if the strings are in the language it will definitely accept it . Now if it accept 2022 string then we don’t need to check as it need to accept at least 2022 string. By this way after finite amount of time TM can confirm for YES cases . So it is turing recognizable.
Good Explanation by arjun sir https://gateoverflow.in/66/consider-the-following-languages
must read:- https://gatecse.in/rices-theorem/
Good playlist:- https://www.youtube.com/watch?v=Pt6GBVIifZA&list=PL7HjUNIdk93ThXvz2Oa_g30Jt3Owwm4HZ
