How to solve the problem:
$\left(3\right)$ \begin{cases} u_{tt}-a^{2}u_{xx}=f\left(x,t\right)\\ u_{x}\left(0,t\right)-h_{0}u\left(0,t\right)=g_{0}\left(t\right)\\ u_{x}\left(L,t\right)-h_{L}u\left(L,t\right)=g_{L}\left(t\right)\\ u\left(x,0\right)=u_{0}\left(x\right) & u_{t}\left(x,0\right)=u_{1}\left(x\right) \end{cases}
My opinion is solve the two following problems before solve $\left(3\right)$
$\left(1\right)$ \begin{cases} u_{tt}-a^{2}u_{xx}=0 & 0<x<L,t>0\\ u_{x}\left(0,t\right)-h_{0}u\left(0,t\right)=u_{x}\left(L,0\right)+h_{L}u\left(L,t\right)=0\\ u\left(x,0\right)=u_{0}\left(x\right) & u_{t}\left(x,0\right)=u_{1}\left(x\right) \end{cases}
$\left(2\right)$ \begin{cases} u_{tt}-a^{2}u_{xx}=f\left(x,t\right)\\ u_{x}\left(0,t\right)-h_{0}u\left(0,t\right)=u_{x}\left(L,0\right)+h_{L}u\left(L,t\right)=0\\ u\left(x,0\right)=0 & u_{t}\left(x,0\right)=0 \end{cases}
One can reduce the general case to the problem with homogeneous boundary conditions $$ \begin{cases} u_{tt}-a^{2}u_{xx}=f\left(x,t\right)\\ u_{x}\left(0,t\right)-h_{0}u\left(0,t\right)=0\\ u_{x}\left(L,t\right)-h_{L}u\left(L,t\right)=0\\ u\left(x,0\right)=u_{0}\left(x\right) & u_{t}\left(x,0\right)=u_{1}\left(x\right) \end{cases} \qquad\qquad(*) $$ by subtracting a (smooth enough) function $v$, satisfying the boundary conditions. And then solving by the Fourier method systems $(1)$ and $(2)$.
For example, if the boundary data is smooth enough, one can search $v$ as a linear function of $x$, $v=a(t)x+b(t)$, satisfying $$ \begin{cases} v_{x}\left(0,t\right)-h_{0}v\left(0,t\right)=g_{0}\left(t\right),\\ v_{x}\left(L,t\right)-h_{L}v\left(L,t\right)=g_{L}\left(t\right).\\ \end{cases} $$ Which gives $$v=\frac{g_0(t) h_L-h_0 g_L(t)}{h_0 L h_L+h_L-h_0}x+\frac{g_0(t)-L g_0(t) h_L-g_L(t)}{h_0 L h_L+h_L-h_0} $$ and $\tilde u=u-v$ satisfies (*) with some $\tilde f$ and $\tilde u_0$, $\tilde u_1$ instead of $f$, $ u_0$, $u_1$.