Relation between Undecidable problems and NP-Hard
I believe that this answer by Yuval Filmus all the questions you have asked.


  If P=NP then any non-trivial set is NP-hard (other than the empty set and the complete set), so assume P$\neq$NP. If $A$ is a set and $f_i$ reduces SAT to $A$ in polytime, then $f_i$ must have infinite range. Otherwise, we can hardcode the relevant values of $f_i$ to get a polytime algorithm for SAT.
  
  We can construct an undecidable problem which is not NP-hard using diagonalization. Let $f_i$ be an enumeration of all polytime reductions whose range is infinite. We construct an undecidable problem $A$ such that no $f_i$ reduces SAT to $A$. We will use $K$ to denote the undecidable set corresponding to the halting problem.
  
  The set $A$ will be defined in stages, starting with a completely undefined set. In stage $i$, we find a string $s$ such that $f_i(s)$ is longer than any string on which $A$ is defined (here we use the fact that the range of $f_i$ is infinite). We define $A$ on $f_i(s)$ so that $s \in SAT$ iff $f_i(s) \notin A$. After all finite stages, we complete the definition of $A$ for each undefined string $s$ by letting $s \in A$ iff $|s| \in K$.
  
  By construction, no polytime $f_i$ reduces SAT to $A$, and so $A$ is not NP-hard. On the other hand, $A$ is not decidable since $K$ reduces to $A$: we can decide whether $n \in K$ (for $n \geq 2$) by taking a majority of three strings of length $n$.


To summarize,


Halting problem is NP-hard.
If $P\ne NP$, not all undecidable problems are NP-hard.
If $P = NP$, all non-trivial sets are NP-hard.


The original answer had not addressed the last part of your question, namely, are there problems which are neither NP nor NP Hard? I will be lazy again and quote another answer, this time by Peter Shor.


  There is a problem which is both NP-hard and in coNP if and only if NP = coNP.
  
  If NP = coNP, than NP-complete problems (like 3-SAT) are both NP-hard and in coNP. 
  
  On the other hand, if any NP-hard problem is in coNP, then all problems in NP are reducible to it, so all problems in NP are in coNP so NP &sube; coNP. Now, since the complement of NP is coNP, and vice versa, we also have  coNP &sube; NP. This means NP = coNP. 
  
  The question of whether NP = coNP is open, but most theoretical computer scientists do not think it is very likely.


So, assuming $NP \ne coNP$, there exist problems that are decidable but neither in NP nor NP-hard. Note that we don't know that $NP = coNP$ implies $P = NP$. So this is a stronger assumption than the one you had suggested ($P \ne NP$).