Given:
$$\frac{\partial F}{\partial t} = \alpha^2 \, \frac{\partial^2 F}{\partial x^2}-h \, F$$
$$F(x,0) = 0, \hspace{5mm} F(0,t) = F(L,t)=F_{0} \, e^{-ht}.$$
The process to obtain a solution is the following.
The boundary conditions suggest making the substitution $F(x,t) = \phi(x,t) \, e^{-h t}$ for which the pde becomes
$$\frac{\partial \phi}{\partial t} = \alpha^{2} \, \frac{\partial^{2} \phi}{\partial x^{2}}$$
where $\phi(x,0) = 0$, $\phi(0,t) = \phi(L,t) = F_{0}$.
Now let $\phi(x,t) = F_{0} + \theta(x,t)$ which bring the equation and conditions into the form
$$\frac{\partial \theta}{\partial t} = \alpha^{2} \, \frac{\partial^{2} \theta}{\partial x^{2}}$$
where $\phi(x,0) = - F_{0}$, $\phi(0,t) = \phi(L,t) = 0$.
Let $\theta(x,t) = f(x) \, g(t)$ to obtain
\begin{align}
\frac{g'}{g} = - \lambda^{2} = \alpha^{2} \, \frac{f''}{f}
\end{align}
for which
\begin{align}
& \alpha^{2} \, f'' + \lambda^{2} f = 0 \\
& g' + \lambda^{2} \, g = 0.
\end{align}
The first order equation has the solution $g(t) = e^{- \lambda^{2} \, t}$. The equation for $f$ has the form $f'' + (\lambda/\alpha)^{2} \, f=0$ with solutions $f(x) = A \, \cos(\lambda x/\alpha) + B \, \sin(\lambda x/\alpha)$.
From the conditions $\phi(0,t)=\phi(L,t) = 0$ then
\begin{align}
0 &= A \\
0 &= A \, \cos\left(\frac{\lambda L}{\alpha}\right) + B \, \sin\left( \frac{\lambda L}{\alpha}\right)
\end{align}
for which $B \neq 0$ and $\sin\left(\frac{\lambda L}{\alpha}\right) = 0$. From this
$$\lambda_{n} = \frac{n \, \pi \, \alpha}{L}.$$
Combining the parts leads to the $\theta(x,t)$ solution
\begin{align}
\theta(x,t) = \sum_{n=1}^{\infty} B_{n} \, \sin\left(\frac{n \, \pi \, x}{L}\right) \, e^{- \frac{n^{2} \, \pi^{2} \, t}{L^{2}}
}\end{align}
The remaining condition is $\theta(x,0)= - F_{0}$,
\begin{align}
- F_{0} = \sum_{n=1}^{\infty} B_{n} \, \sin\left(\frac{n \, \pi \, x}{L}\right)
\end{align}
The coefficients are obtained by Fourier series methods and are
\begin{align}
B_{m} = - \frac{2}{L} \, \int_{0}^{L} F_{0} \, \sin\left(\frac{m \, \pi \, u}{L}\right) \, du = - \frac{2 \, F_{0} \, (1 - (-1)^{m})}{m \, \pi}.
\end{align}
With all of this the solution becomes
\begin{align}
F(x,t) = F_{0} \, e^{-h t} - \frac{2 \, F_{0}}{\pi} \, e^{-h t} \, \sum_{n=1}^{\infty} \frac{1 - (-1)^{n}}{n} \, \sin\left(\frac{n \, \pi \, x}{L}\right) \, e^{- \frac{n^{2} \, \pi^{2} \, t}{L^{2}}}.
\end{align}
or
\begin{align}
F(x,t) = F_{0} \, e^{-h t} - \frac{2 \, F_{0}}{\pi} \, e^{-h t} \, \sum_{n=0}^{\infty} \frac{1}{2n+1} \, \sin\left(\frac{(2n+1) \, \pi \, x}{L}\right) \, e^{- \frac{(2n+1)^{2} \, \pi^{2} \, t}{L^{2}}}.
\end{align}