Prove (non-)recursiveness of a set of Turing Machines defined by behavior
$\boxed{\mathcal R}$

Idea: If you never move right, you have a regular behaviour. And you can move right iff a "state" in which you can move right is reachable while following this regular behaviour.

Solution:


   $\mathcal R$ is decidable


Proof:


   Let $\mathcal M$ be a TM with states $Q$ and tape alphabet $\Gamma$. Build the following automaton:

 - States: $Q\times (\Gamma \sqcup \{?\})$ where the right component is whatever is on the tape at the current position (we write $?$ if it is whatever was there at the very beginning, and $a$ if we wrote an $a$ there)

 - Transitions:
 $\hspace{0.5em}$ - $(p, a)\overset{a}{\to}(q,?)$ if the TM can read $a$ while in state $p$ and move left
 $\hspace{0.5em}$ - $(p, a)\overset{\varepsilon}{\to}(q,b)$ if the MT can write a $b$ and move to state $q$ if it's in state $p$ and reads an $a$
 $\hspace{0.5em}$ - $(p, ?)\overset{\varepsilon}{\to}(p,b)$ (because since we just moved left and never moved right, replacing the letter at that position on the tape by anything wouldn't change the execution until that point).
 - The initial states are $(q_i, ?)$ where $q_i$ is an initial state of the TM.

 Then in this automaton, a state $(p,a)$ is accessible iff there is a run in the TM so that gets in a state where the TM is in state $p$ and is about to read an $a$ on the tape. So you just have to check if there is an accessible state $(p,a)$ in this automaton so that the TM can read an $a$ while in state $p$ and move right.




$\boxed{\mathcal S}$

Idea: Halting for all inputs is "For all input, there is an execution that halts" which is more complicated than the halting problem "Given this input, there is an execution that halts".

Solution:


   $\mathcal S$ is undecidable


Proof:


   Take a TM $\mathcal M$ and an input $w$. Build a new TM $\mathcal M'$ that erases its input, writes $w$ and then runs $\mathcal M$.

 $\mathcal M'$ halts on all inputs
 (iff $\mathcal M'$ halts on some input)
 iff $\mathcal M$ halts on input $w$

 So your problem is undecidable because the halting problem can be reduced to it.