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?

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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: 
This is a partial answer that shows an obstruction to certain eigenvalue sequences. First, I claim that if $M$ is rank one then it isn't similar to something of the stated form. 

Take a matrix $M$ of the given form and suppose the rank is $0$ or $1.$ If $a\neq0$, then the rank of $M$ is at least 4. If $b\neq0$ or $c\neq0$ then the rank of $M$ is at least 2. Since $M$ is rank 0 or 1 we must have $a=b=c=0.$ Since we have a 0 diagonal, if $A\neq0$, then the rank of $M$ would have to be at least 2 so we must have $A=0$. Similarly, the 0 diagonal and $B\neq0$ forces the rank to be at least 2 so we must have $B=0$, so $M$ is rank $0.$

That eliminates some eigenvalue sequences. Now notice that if the rank of $M$ is $\leq 3$ and $M$ is positive semidefinite this forces $a=0$ and $A=0$ so you are back in the case that you know how to deal with. This will eliminate some other eigenvalue sequences.
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.