Question: Edit: since we seem a bit deadlocked at this point, let me weaken the question. It's fairly easy to see that the set of 8-tuples of reals which can be the eigenvalues of a matrix of the desired form is closed. We know from jjcale and Caleb Eckhardt that its complement is nonempty. Is its complement dense? That is, would a generic 8-tuple not be the eigenvalues of such a matrix?

---

First, here is a baby version of the question, that I already know the answer to. Consider complex Hermitian $4\times 4$ matrices of the form $$\left[\begin{matrix}a I\_2&A\cr A^\*&b I\_2\end{matrix}\right]$$ where $A \in M\_2(\mathbb{C})$ and $a,b \in \mathbb{R}$ are arbitrary. Can any four real numbers $\lambda\_1 \leq \lambda\_2 \leq \lambda\_3\leq \lambda\_4$ be the eigenvalues of such a matrix, or is there some restriction? Answer: there is a restriction, we must have $\lambda\_1 + \lambda\_4 = \lambda\_2 + \lambda\_3$.

The real question is: what are the possible eigenvalues of Hermitian $8\times 8$ matrices of the form $$\left[\begin{array}{c|c}aI\_4&A\cr \hline A^\*&\begin{matrix}bI\_2& B\cr B^\*&cI\_2\end{matrix}\end{array}\right]$$ with $a,b,c\in\mathbb{R}$, $A \in M\_4(\mathbb{C})$, and $B \in M\_2(\mathbb{C})$? Can any eight real numbers be the eigenvalues of such a matrix? (I suspect not. If they could, that would tell you that any Hermitian $8\times 8$ matrix is unitarily equivalent to one of this form.)

Answer: 
Let $V$ be the real vector space of the $8\times 8$ matrices of the form given in the question.

Where is an open set in $\mathbb{R}^8$ of possible $8$-tuples of eigenvalues of matrices in $V$.

**Proof :**

Choose $M\_1 \in V$ such that $M\_1$ has no degenerated eigenvalues.

Let $v\_1,...,v\_8$ be an orthonormal base of eigenvectors of $M\_1$ .

Then first order perturbation theory tells us that it suffices to show that the map

$V \rightarrow \mathbb{R}^8$, $M \mapsto (v\_1^\* M v\_1,...,v\_8^\* M v\_8)$ has rank $8$.

So I need $7$ additional matrices $M\_2,...,M\_8$ in $V$ such that the matrix
$X = (v\_i^\* M\_j v\_i)\_{ij}$ is nonsingular . 

Let $f(a,b,c,A,B)$ be the corresponding matrix.

Choose $$ A\_1 = \left[\begin{matrix}4&2&3&4\cr5&6&7&8\cr9&10&11&12\cr13&14&15&16\end{matrix}\right] $$ .

Then choose

$M\_1 = f(2.7,1,-1,A\_1,diag(2,1))$,

$M\_2 = f(1,0,0,0,0)$,

$M\_3 = f(0,1,0,0,0)$,

$M\_4 = f(0,0,1,0,0)$,

$M\_5 = f(0,0,0,diag(1,1,1,2),0)$,

$M\_6 = f(0,0,0,diag(1,0,0,0),diag(0,1))$,

$M\_7 = f(0,0,0,diag(0,1,0,0),diag(1,2))$,

$M\_8 = f(0,0,0,diag(0,0,1,0),diag(3,2))$.

This gives $det X = 21.661...$ .
It should follow from Cauchy's interlacing theorem (with compression $\begin{bmatrix}I\_2\\0\end{bmatrix}$) that your first example has two eigenvalues greater than $a$ and two smaller. The same holds for $b$, considering $\begin{bmatrix}0\\I\_2 \end{bmatrix}$ as compression; so in practice you have 2 eigenvalues larger than $\max(a,b)$ and two smaller than $\min(a,b)$.

Similarly, in your second example you have four eigenvalues greater than $a$ and four smaller. This argument doesn't tell all the story (for instance, in your first example it does not reveal the constraint on the sums that you have found), but at least it shows in a more general setting that there are some restrictions on the eigenvalues.