Can a field extension still have "non-separability" above its maximal purely inseparable subextension? Question 1
Let $E/F$ be an algebraic field extension. Let $K$ be the set of all elements of $E$ that are purely inseparable over $F$. Then, $E/K/F$ is a tower of fields, and $K/F$ is purely inseparable.
In this case, is $E/K$ always separable? If not, what is a counterexample?
Moreover, let $K^s$ and $F^s$ be the separable closures in $E$. Then, $E/K^s$ is purely inseparable. How different are $K^s$ and $F^s$ as fields? That is, how much nicer is the extension $K^s/K$ compared to $F^s/F$?
Question 2
Let $E/F$ be a normal field extension. Let $E^G$ denote the fixed field of $\operatorname{Aut}(E/F)$. Then, $E/E^G$ is separable and $E^G/F$ is purely inseperable.
Is $E^G$ the unique such subextension? That is, is there a subextension $K$ of $F$ such that $E/K$ is separable and $K/F$ is purely inseparable, but $K\neq E^G$?
 A: The extension $E/K$ need not be separable. Here is the example I learned from a note by J. Lipman.
Consider the rational function field $F=\mathbb{F}_2(y,z)$ and the extension $E=F(x)$, where $x$ is a root of $$f(t)=t^4+yt^2+z\in F[t].$$ If $E/K$ was separable, we would have $f=g^2$, for $g\in K[t]$. We have $g=t^2+\sqrt{y}t+\sqrt{z}$, which means that $\sqrt{x},\sqrt{y}\in K$. The latter condition says that $K/F$ is a degree-four extension, and so it is not purely inseparable.
A: Answer to Question 1.
Here is another example of an extension $K/F$ such that $K/I$ is not separable, where $I$ is the purely inseparable closure of $F$ in $K$. This is Example 4.24 from Patrick Morandi's Field and Galois Theory (pages 48-49), and it is similar in spirit to the one mentioned in @user2097's answer.
Let $k$ be a field of characteristic $2$. Let $F = k(x,y)$ where $x$ and $y$ are indeterminates. Let $S = F(u)$, where $u$ is a root of the polynomial
$$
t^2 + t + x \in F[t].
$$
Let $K = S(\sqrt{uy})$. Then $K/S$ is purely inseparable and $S/F$ is separable, so $S$ is the separable closure of $F$ in $K$. But the purely inseparable closure of $F$ in $K$ equals $F$ itself, and so $K$ is not separable over $I$.
I cannot provide a strong answer for how much nicer $I^s/I$ is compared to $F^s/F$. Perhaps you would receive better answers if you asked it separately, either here or on MathOverflow.

Answer to Question 2.
Note that for any algebraic extension $E/F$, we have that $E/E^G$ is separable and that $E^G \supset I$, and if $E/F$ is normal, then $E^G = I$.
Now, let $E/F$ be a normal extension, and let $K$ be an intermediate extension such that $E/K$ is separable and $K/F$ is purely inseparable. Then, we have the chain of extensions
$$
F \subset K \subset I \subset E.
$$
Since $E/K$ is separable and $E/I = E/E^G$ is separable, $I/K$ is also separable. But $I/F$ is purely inseparable, so $I/K$ is also purely inseparable. Hence, $I = K$. Thus, if $E/F$ is normal, then there is a unique intermediate extension such that $E/F$ splits into a purely inseparable extension followed by a separable extension, namely $I = E^G$.
If $E/F$ is not necessarily normal, then there may not exist such an intermediate extension at all. If it does exist, then that intermediate extension is unique and equals $I$, but $E^G$ need not equal $I$ in this case! This follows from the results in the note of J. Lipman that @user2097 mentioned in their answer.
