Incomputable sets of low degree vs Rices theorem? I have heard that there are sets that are not computable, but are lower in degree than the halting problem.
How does this not contradict Rice's theorem? Are there any concrete examples of such sets?
This is not an answer, but way too long to be a comment.  I just wanted to cite Robert I. Soare's Recursively Enumerable Sets and Degrees, Chapter V: Simple Sets and Post's Problem.

1: Immune Sets, Simple Sets, and Post's Construction
The only r.e. degrees constructed so far are $\textbf{0}$ and
$\textbf{0}'$.  Post's Problem was to construct other r.e. degrees,
i.e., to construct an r.e. set $A$ such that
$\emptyset \lt_\text{T} A \lt_\text{T} \emptyset'$.  Post's Program
for constructing such a set $A$ was to ﬁnd some easily deﬁnable property
on $\overline{A}$ (compatible with $A$ being nonrecursive) which
guarantees incompleteness ($K \not\leq_\text{T} A$).
Recall (from Theorem II.4.5) that $K$ (and hence any creative set) has
an inﬁnite r.e. set (and therefore inﬁnitely many inﬁnite r.e. sets)
contained in its complement. Post's idea for constructing $A$ incomplete
was to make $\overline{A}$ sufﬁciently &quot;thin&quot; with respect to
containment of inﬁnite r.e. sets so that $\overline{K}$ could not be
Turing reduced to $\overline{A}$.
1.1 Deﬁnition. (i) A set is immune if it is infinite but contains no
infinite r.e. set.
(ii) A set $A$ is simple if $A$ is r.e. and $\overline{A}$ is immune.
1.2 Proposition. If $A$ is simple then:
(i) $A$ is not recursive;
(ii) $A$ is not creative;
(iii) $A$ is not $m$-complete (i.e., $K \not\leq_m A$).
Proof. (i) $\overline{A}$ cannot be r.e., since otherwise it contains an
inﬁnite r.e. set (namely $\overline{A}$).
(ii) The complement of a creative set does contain an inﬁnite r.e. set (by
Theorem II.4.5).
(iii) If $K \leq_m A$, then $A$ is creative (by Corollary II.4.8). $\Box$
1.3 Theorem (Post [1944]). There exists a simple set $S$.
Proof. Let $A = \{(e,x) \;:\; x \in W_e \;\&amp;\; x &gt; 2e \}$.  Now $A$ is
$\Sigma_1$ and hence r.e. Let $\psi$ be any p.r. selector function for
$A$ (in the sense of Theorem II.1.6) and let $S = \text{rng}\; \psi$.
(Intuitively, enumerate $W_e$ until the ﬁrst element $\psi(e) &gt; 2e$
appears in $W_e$, and then put $\psi(e)$ into $S$.) The following facts
give the simplicity of $S$.
(1) $S$ is r.e. ($S$ is the range or a p.r. function.)
(2) $\overline{S}$ is inﬁnite.  To see this, note that $S$ contains at
most $e$ elements out of $\{0,1,2,\dots,2e\}$, namely $\psi(0),\psi(1), \dots, \psi(e-1)$. Hence $e + 1 \leq \text{card}(\overline{S} \upharpoonright (2e+1))$, so $\overline{S}$ is inﬁnite.
(3) If $W_e$, is inﬁnite, then $W_e \cap S \neq \emptyset$, because
$(e,x) \in A$ for some $x &gt; 2e$ and so $\psi(e)$ is deﬁned and $\psi(e) \in S \cap W_e$. $\Box$