There are twos cases to consider:
$\mathbf{1.}\,H=0$: In this case, we have to solve
$$
f''+\lambda f=0, \tag{1}
$$
subject to the boundary conditions
$$
f(0)=f(a)=0 \tag{2}
$$
and to the condition
$$
\int_0^{a}f''(x)\,dx=f'(a)-f'(0)=0. \tag{3}
$$
It is well known that the nontrivial solutions to $(1)$ satisfying $(2)$ are $f_n(x)=C_n\sin(\sqrt{\lambda_n}x),$ where $C_n\neq 0$ and $\lambda_n=\left(\frac{n\pi}{a}\right)^2$ $(n\in\mathbb{N}^*).$
However, not all of these solutions satisfy $(3)$. Indeed,
\begin{align}
f_n'(a)-f_n'(0)=0 &\implies C_n\left(\frac{n\pi}{a}\right)(\cos(n\pi)-1)=0 \\
&\implies n=2m\quad(m\in\mathbb{N}^*).\phantom{\left(\frac{n\pi}{a}\right)} \tag{5}
\end{align}
Therefore, there are infinitely many eigenfunctions and eigenvalues in the case $H=0$, given by
$$
f_m(x)=C_m\sin(\sqrt{\lambda_m} x) \quad\text{and}\quad \lambda_m=\left(\frac{2m\pi}{a}\right)^2\quad(m\in\mathbb{N}^*). \tag{6}
$$
$\mathbf{2.}\,H\neq 0$: Now we have to solve the differential equation
$$
f''+\lambda f=H \tag{7}
$$
subject to $(2)$ and to the consistency condition
$$
\frac{1}{a}\int_0^a f''(x)\,dx=\frac{f'(a)-f'(0)}{a}=H. \tag{8}
$$
Depending on the sign of $\lambda$, the solution to $(7)$ satisfying $(2)$ is
$$
f(x)=\begin{cases}
\frac{H(\sinh q(a-x)+\sinh qx-\sinh qa)}{q^2\sinh qa}&\text{if $\lambda=-q^2<0$,} \\
\frac{1}{2}Hx(x-a)&\text{if $\lambda=0$,} \\
-\frac{H(\sin k(a-x)+\sin kx-\sin ka)}{k^2\sin ka}&\text{if $\lambda=k^2>0\quad(ka\neq n\pi, n\in\mathbb{N}^*).$}
\end{cases} \tag{9}
$$
Let's now plug into $(8)$ each of the solutions in $(9)$:
For $\lambda=-q^2<0$, we get
\begin{align}
H&=\frac{2H(-1+\cosh qa)}{qa\sinh qa}=\frac{2H}{qa}\tanh\left(\frac{qa}{2}\right) \\
&\qquad\implies \tanh\left(\frac{qa}{2}\right)=\frac{qa}{2}. \tag{10}
\end{align}
The only real solution to Eq. $(10)$ is $q=0$, but it must be discarded because we are assuming $q\neq 0$. Therefore, there are no negative eigenvalues.
For the solution corresponding to $\lambda=0$, $(8)$ yields the identity $H=H$. This shows that $\lambda=0$ is an eigenvalue, and its associated eigenfunction is the second solution in $(9)$.
Finally, for $\lambda=k^2>0$ we get
\begin{align}
H&=\frac{2H(1-\cos ka)}{ka\sin ka}=\frac{2H}{ka}\tan\left(\frac{ka}{2}\right) \\
&\qquad\implies \tan\left(\frac{ka}{2}\right)=\frac{ka}{2}.
\tag{11}
\end{align}
Equation $(11)$ has infinitely many positive solutions$^{(*)}$. This gives us another infinite set of eigenfunctions and eigenvalues.
$^{(*)}$ One can find approximate solutions to the equation $\tan x=x$ in this post.
