Here were my solutions from a few years ago, when I worked through Jech's 1984 paper on this topic.

**Exercise 8.13**:
If $\lambda<\kappa$ is the $\alpha$th regular cardinal, then $o(E_{\lambda}^{\kappa})=\alpha$.

*Proof.*
Now, $o(E_{\lambda}^{\kappa})\geq \alpha$ follows from the fact that $E_{\nu}^{\kappa}<E_{\mu}^{\kappa}$ when $\nu<\mu<\kappa$ are regular cardinals. (This later fact follows from $E_{\mu}^{\kappa}\subset {\rm Tr}(E_{\nu}^{\kappa})$, by the previous exercise.)

The other inequality is by induction, and we follow the proof given in Jech [1984: Stationary Subsets of Inaccessible Cardinals]. (We will show that any stationary set of order $\alpha$ must have an element of cofinality the $\alpha$th regular cardinal.) All of our stationary sets will be restricted to limit ordinals.

For each ordinal $\alpha<\kappa^{+}$, we say that a stationary set $E$ is *canonical* of order $\alpha$ if (1) $o(X)=\alpha$ for every stationary set $X\subset E$, and (2) $E$ meets every stationary set of order $\alpha$. Such a canonical set, if it exists, is clearly unique (modulo nonsingular sets). If it exists, we write $E_{\alpha}$ for such a set. (We will eventually show $E_{\alpha}=E_{\lambda}^{\kappa}$.)

**Lemma 3.1 from Jech's paper:** For a stationary set $S$, if $o(S)>\alpha$ and $E_{\alpha}$ exists, then $S\cap {\rm Tr}(E_{\alpha})\neq \emptyset$.

*Proof of Lemma 3.1.*
The proof is by contradiction, letting $S$ be a minimal counterexample. Let $T<S$ be stationary of order $\nu$. Write $T=A\cup B\cup C$ where $A=T\cap E_{\alpha}$, $B=T\cap {\rm Tr}(E_{\alpha})$, and $C=T-(A\cup B)$. Since $A\subset E_{\alpha}$ we have ${\rm Tr}(A)\subset {\rm Tr}(E_{\alpha})$, so ${\rm Tr}(A)$ is disjoint from $S$. Next, $B\subset {\rm Tr}(E_{\alpha})$, hence ${\rm Tr}(B)\subset {\rm Tr}({\rm Tr}(E_{\alpha}))\subset {\rm Tr}(E_{\alpha})$ is also disjoint from $S$. From $T<S$ we have $S\subset {\rm Tr}(C)\mod I_{\rm NS}$. In particular, $C$ is stationary and $C<S$. Since taking the order reverses inclusion, we have $o(C)\geq o(T)=\alpha$. From heredity, we cannot have $o(C)=\alpha$, so the inequality is strict. But then we have a contradiction on the minimality of $S$. $\square_{\rm Lemma\ 3.1}$

**Corollary 3.2 from Jech's paper:** ${\rm Tr}(E_{\alpha})$ is the largest set (modulo nonstationary sets) of order $>\alpha$, if it is non-empty modulo $I_{\rm NS}$. In other words, for any stationary set $S$, if $o(S)>\alpha$ then $S>E_{\alpha}$.

*Proof of Corollary 3.2.*
Assume not, so there is some stationary set $S$ with $o(S)>\alpha$ but $S$ is not contained in ${\rm Tr}(E_{\alpha})$ (modulo $I_{\rm NS}$). Let $T:=S-{\rm Tr}(E_{\alpha})$, which is a stationary set by the assumption. Further, $T$ is disjoint from ${\rm Tr}(E_{\alpha})$ and $o(T)\geq o(S)$ (from order reversal from $T\subset S$), a contradiction to the previous lemma. $\square_{\rm Corollary\ 3.2}$

**Lemma 3.3 from Jech's paper:** The set $E_{\alpha}$ exists if and only if there exists a largest stationary set $M_{\alpha}$ of order $\alpha$. Then
\begin{equation}\tag{A}\label{Eq:3.3fromJech}
M_{\alpha}=E_{\alpha}\cup {\rm Tr}(E_{\alpha})
\end{equation}
and
\begin{equation}\tag{B}\label{Eq:3.4fromJech}
E_{\alpha}=M_{\alpha}-{\rm Tr}(M_{\alpha}).
\end{equation}
The union in (\ref{Eq:3.3fromJech}) is disjoint, and ${\rm Tr}(M_{\alpha})\subset M_{\alpha}$. (Both statements are taken modulo $I_{\rm NS}$.) Moreover, ${\rm Tr}(E_{\alpha})={\rm Tr}(M_{\alpha})$.

*Proof of Lemma 3.3.*
$(\Rightarrow)$: Assume $E_{\alpha}$ exists. Note that $E_{\alpha}$ and ${\rm Tr}(E_{\alpha})$ are disjoint modulo $I_{\rm NS}$ since $o({\rm Tr}(E_{\alpha}))>\alpha$ and the heredity property on $E_{\alpha}$. Putting $M:=E_{\alpha}\cup {\rm Tr}(E_{\alpha})$, the order of $M$ is $\alpha$. [Prove by induction, using minimal counterexample, that $o(S\cup T)=\min\{o(S),o(T)\}$ for stationary sets. We call this the union property.] If $o(S)=\alpha$ write $S=A\cup B$ where $A=S\cap E_{\alpha}$ and $B=S-A$. If $B$ is nonstationary, then $S=A\subset M$ modulo $I_{\rm NS}$. If $B$ is stationary, by the second defining property on $E_{\alpha}$ and the union property we have $o(B)>\alpha$, and so $B\subset {\rm Tr}(E_{\alpha})\mod I_{\rm NS}$ by the previous corollary. Hence in this case we also have $S\subset M\mod I_{\rm NS}$; proving that $M$ is the largest set of order $\alpha$.

$(\Leftarrow)$: Assume $M_{\alpha}$ exists. By the union property [or by trivial considerations when ${\rm Tr}(M_{\alpha})$ is nonstationary] it follows that $M_{\alpha}\cup {\rm Tr}(M_{\alpha})$ has order $\alpha$, so ${\rm Tr}(M_{\alpha})\subset M_{\alpha}$ (modulo $I_{\rm NS}$) by maximality. Let $E=M_{\alpha}-{\rm Tr}(M_{\alpha})$. The order of $E$ is $\alpha$ since $M_{\alpha}=E\cup {\rm Tr}(M_{\alpha})$, by the union property, etc.

We first show that $E$ has order $\alpha$ hereditarily. Let $S\subset E$ be stationary. If $o(S)>\alpha$ then there is a stationary set $T<S$ of order $\alpha$. By maximality of $M_{\alpha}$ we have $T\subset M_{\alpha}\mod I_{\rm NS}$, so $S\subset {\rm Tr}(T)\subset {\rm Tr}(M)\mod I_{\rm NS}$ which contradicts the fact that $S$ does not intersect ${\rm Tr}(M_{\alpha})$.

Finally we show that $E$ meets every stationary set of order $\alpha$. Suppose, by way of contradiction, that $S$ is stationary of order $\alpha$ and $S\cap E=\emptyset$. We have $S\subset M_{\alpha}\mod I_{\rm NS}$, and since $S\cap E=\emptyset$ we have $S\subset {\rm Tr}(M_{\alpha}\mod I_{\rm NS}$, hence $M<S$, contradicting the order. This proves that $E$ is the canonical set of order $\alpha$.

Finally (working modulo $I_{\rm NS}$ everywhere), since $E\subset M$ we have ${\rm Tr}(E)\subset {\rm Tr}(M)$, while the reverse containment follows from the previous corollary.$\square_{\rm Lemma\ 3.3}$

**Corollary 3.4 from Jech's paper:** Assuming $M_{\alpha}$ exists, for any stationary set $S$, then $o(S)\geq \alpha$ if and only if $S\subset M_{\alpha}\mod I_{\rm NS}$.

*Proof of Corollary 3.4.*
This is just a direct application of the definition of $M_{\alpha}$ and the union property (taking the union with $M_{\alpha}$). [I don't see how it is a corollary.]
$\square_{\rm Corollary\ 3.4}$

**Lemma 3.5 from Jech's paper:** [Note, it is misnumbered there.] Let $\alpha$ be a limit ordinal and assume that $M_{\beta}$ exists for all $\beta<\alpha$. Then $M_{\alpha}$ exists if and only if ${\rm inf}_{\beta<\alpha}M_{\beta}$ exists, and is nonzero, in the Boolean algebra $P(\kappa)/I_{\rm NS}$.

*Proof of Lemma 3.5.*
$(\Rightarrow)$: Assume $M_{\alpha}$ exists. We have $M_{\beta}\supset M_{\alpha}\mod I_{\rm NS}$ when $\beta<\alpha$ by the previous corollary, so $M_{\alpha}$ is a lower bound on the $M_{\beta}$. If $S$ is any stationary set which is a lower bound on each $M_{\beta}$ modulo $I_{\rm NS}$, then again by the previous corollary $o(S)\geq \beta$ for each $\beta<\alpha$, hence $o(S)\geq \alpha$, so $S\subset M_{\alpha}\mod I_{\rm NS}$ as desired.

$(\Leftarrow)$: If the inf exists and is nonzero, call it $M$. As it is contained in each $M_{\beta}$ (for $\beta<\alpha$) it has order at least $\alpha$. Given any stationary subset $S$ with $o(S)=\alpha$ then by the previous corollary $S\subset M_{\beta}\mod I_{\rm NS}$ for each $\beta<\alpha$, hence it lies inside the infimum. In particular, the infimum has exactly order $\alpha$ and is the largest set containing all such sets.$\square_{\rm Lemma\ 3.5}$

**Lemma 3.6 from Jech's paper:** If $M_{\alpha}$ exists and the height of $\kappa$ is at least $\alpha+2$ (i.e.\ there is a set of order $\alpha+1$), then $M_{\alpha+1}={\rm Tr}(M_{\alpha})$.

*Proof of Lemma 3.6.*
We know ${\rm Tr}(S)\geq o(S)+1$ (when ${\rm Tr}(S)$ is stationary). Thus $o({\rm Tr}(M_{\alpha}))\geq \alpha+1$. By Corollary 3.4, ${\rm Tr}(M_{\alpha})\subset M_{\alpha+1}\mod I_{\rm NS}$. Conversely, $o(M_{\alpha+1})>\alpha$ so by Corollary 3.2 $M_{\alpha+1}\subset {\rm Tr}(E_{\alpha})={\rm Tr}(M_{\alpha})\mod I_{\rm NS}$.
$\square_{\rm Lemma\ 3.6}$

As diagonal intersection gives inf for sets of cardinality $<\kappa^{+}$, the previous two lemmas complete an induction, so we see that we can find the maximal sets $M$ inductively for $\alpha<\kappa^{+}$ (when the height of $\kappa$ is bigger than $\alpha$) by: $M_{0}$ is the limit points below $\kappa$, $M_{\alpha+1}={\rm Tr}(M_{\alpha})$, and for limits $\alpha$ we have $M_{\alpha}=\triangle_{\beta<\alpha}M_{\beta}$.

In the first paragraph we showed that the height of $\kappa$ is at least the ordinality of the sequence of regular infinite cardinals $<\kappa$. It is easy to check (via induction) using these definitions that $E_{\alpha}=E_{\lambda}^{\kappa}$ as claimed.$\square_{\rm Exercise\ 8.13}$

**Exercise 8.14:**
The height of $\kappa$ is at least $\kappa$ if and only if $\kappa$ is weakly inaccessibly; the height of $\kappa$ is at least $\kappa+1$ if and only if $\kappa$ is weakly Mahlo.

*Proof.*
First suppose $\kappa$ is an uncountable, regular, successor cardinal. Say $\kappa$ is the $\alpha$th regular cardinal. We have $\alpha<\kappa$. If $\alpha$ is a successor ordinal, say $\alpha=\beta+1$, let $\lambda$ be the $\beta$th regular cardinal. Then $E_{\beta}=E_{\lambda}^{\kappa}$ and ${\rm Tr}(E_{\beta})=\emptyset$ by Exercise 8.12, so the height of $\kappa$ is $\alpha$ by Lemma 3.6 in the solution to the previous problem. If $\alpha$ is a limit ordinal, then since $\triangle_{\beta<\alpha}E_{\alpha}=\triangle_{\lambda<\kappa:\lambda\text{ regular}}E_{\lambda}^{\kappa}=\emptyset$, by Lemma 3.5 in the solution of the previous problem we have the height of $\kappa$ is $\alpha$.

Second suppose $\kappa$ is an uncountable, regular, limit cardinal. This means the sequence of limit cardinals $<\kappa$ is a cofinal sequence, so the height of $\kappa$ is at least $\kappa$, again by the previous problem.

Putting these together, we see $h(\kappa)\geq \kappa$ iff $\kappa$ is weakly inaccessible.

Now, fix $\kappa$ to be weakly inaccessible. The diagonal intersection of the sequence $E_{\lambda}^{\kappa}$ consists of regular cardinals $<\kappa$, so this is stationary if and only if $\kappa$ is weakly Mahlo. Hence by Lemma 3.5 of the previous solution, $h(\kappa)\geq \kappa+1$.

Conversely, if $h(\kappa)\geq \kappa+1$, the diagonal intersection of $E_{\lambda}^{\kappa}$ is stationary, hence there is a stationary sequence of regular cardinals $<\kappa$, so $\kappa$ is weakly Mahlo.$\square_{\rm Exercise\ 8.14}$