$\newcommand{\al}{\alpha}\newcommand\la\lambda\newcommand\R{\mathbb R}$Such a construction of $M$ and $\al$ is always possible.

Indeed, take any complex $\la$. Rearranging columns and rows of the matrix $A-\la I_{3n}$, we see that $\la$ is an eigenvalue of $A$ iff
\begin{equation*}
D(\la):=\begin{vmatrix}
-\la I&I&0\\
0&(1-\la)I&-\al I \\
0&B&C-\la I
\end{vmatrix}=0,
\end{equation*}
where $|\cdot|$ denotes the determinant,
\begin{equation*}
B:=(M+H)^{-1},\quad C:=B(M-2\al I),
\end{equation*}
$I:=I_n$.

Note that $D(\la)$ is the determinant of a block-triangular matrix, so that
\begin{equation*}
D(\la)=(-\la)^n
\begin{vmatrix}
(1-\la)I&-\al I \\
B&C-\la I
\end{vmatrix}.
\end{equation*}
So, $D(1)\ne0$, since $B=(M+H)^{-1}$ is nonsingular.

So, without loss of generality (wlog), $\la\ne1$, and then, by "The general case",
\begin{equation*}
\begin{aligned}
D(\la)&=(-\la)^n(1-\la)^n\,|C-\la I-B((1-\la)I)^{-1}(-\al I)| \\
& =(-\la)^n(1-\la)^n\,|B(M-2\al I)-\la I+\al(1-\la)^{-1}B| \\
& =(-\la)^n(1-\la)^n\,|B|\,|(M-2\al I)-\la(M+H)+\al(1-\la)^{-1}I| \\
& =(-\la)^n(1-\la)^n\,|B|\,d(\la),
\end{aligned}
\end{equation*}
where
\begin{equation*}
d(\la):=|(1-\la)M-\la H+\al((1-\la)^{-1}-2)I|.
\end{equation*}

So, $\la$ is a nonzero eigenvalue of $A$ iff $d(\la)=0$.

By diagonalization, wlog the matrix $H$ is diagonal, with (say) real $h_1,\dots,h_n$ on its diagonal. Letting now $M$ be diagonal as well, with positive real $m_1,\dots,m_n$ on its diagonal, we see that
\begin{equation*}
d(\la)=\prod_{i=1}^n f_{\al,h_i}(\la,m_i),
\end{equation*}
where $f_{\al,h}(\la,m):=(1-\la)m-\la h+\al((1-\la)^{-1}-2)$.

For $\la\ne1$, the equation $f_{\al,h}(\la,m)=0$ for $\la$ is equivalent to a quadratic equation, with roots
\begin{equation}
\la_\pm:=\la_\pm(\al,h,m):=\frac{h+2 m-2 \al \pm\sqrt{4 \alpha ^2+h^2-4 \al m}}{2 (h+m)}.
\end{equation}
Taking now any $\al\in(\max(0,-h),\infty)$ and then choosing $m=\frac{4\al^2+h^2}{4\al}$, we get
$\la_+=\la_-=\frac h{2\al+h}\in(-1,1)$.

So, for any real $\al>\max(0,-h_1,\dots,-h_n)$ we can find positive real $m_1,\dots,m_n$ such that all the roots $\la$ of the equation $d(\la)=0$ are in the interval $(-1,1)$.

Thus, we will have all the eigenvalues of $A$ in the interval $(-1,1)$. $\quad\Box$