I came across this question while doing a problem set with a similar exercise, so hopefully this answer is a helpful reference.
You made a few typos after you derived the tangent approximation, where you switched the ordering and sign of the $\beta$ terms, used the $\tau$ for $x$ at $x+\Delta x$, and changed to taking $x$-derivatives on the right hand side.
I fix these below:
Plug these back into the equation with the sines of alpha and beta
$-\tau \tan \alpha + \tau \tan \beta = \left[\rho \frac{\partial ^2 u}{\partial t^2} \Delta x \right] \Rightarrow \left[-\tau(x) \frac{\partial u}{\partial x}(x,t) + \tau(x+\Delta x)\frac{\partial u}{\partial x}(x+\Delta x, t)\right] = \rho \frac{\partial ^2 u}{\partial t^2} \Delta x$
We divide both sides by $\Delta x$:
$$\frac{\left[-\tau(x) \frac{\partial u}{\partial x}(x,t) + \tau(x+\Delta x)\frac{\partial u}{\partial x}(x+\Delta x,t)\right]}{\Delta x} = \rho \frac{\partial ^2 u}{\partial t^2} $$
Now notice that for any function $f$, $$\lim_{\Delta\to 0} \frac{\left[ -f(x) + f(x + \Delta x) \right]}{\Delta x} = \frac{\partial f}{\partial x}(x)$$, by definition of the derivative.
So we take the limit of our last step and substitute this definition in:
$$\frac{\partial\left[ \tau \frac{\partial u}{\partial x}\right]}{\partial x} = \rho \frac{\partial ^2 u}{\partial t^2},$$
which is as required.
Because you should be expecting oscillatory behavior in $x$, maybe it would be convenient to call the constant $-\lambda$. Then the $x$ equation becomes
$$X''+\frac{\lambda}{c^2} X=0$$
so that
$$X(x) = A \cos{\left ( \frac{\sqrt{\lambda}}{c} x\right )} + B \sin{\left ( \frac{\sqrt{\lambda}}{c} x\right )}$$
Because the string is fixed at the ends, the boundary conditions are
$$X(0) = X(L) = 0$$
$X(0)=0 \implies A=0$, and $X(L)=0$ means that
$$\sin{\left ( \frac{\sqrt{\lambda}}{c} L\right )}=0$$
This only happens for certain values of the argument, i.e., when
$$\frac{\sqrt{\lambda}}{c} L = n \pi \implies \lambda_n = \left (\frac{n \pi c}{L} \right )^2 $$
The $t$ equation becomes
$$T'' + \left [ \gamma^2 + \left (\frac{n \pi c}{L} \right )^2\right ] T = 0$$
which means that
$$T_n(t) = C_n \cos{\eta_n t} + D_n \sin{\eta_n t}$$
where
$$\eta_n = \sqrt{\gamma^2 + \left (\frac{n \pi c}{L} \right )^2}$$
The general solution is a linear combination over each $n$:
$$u(x,t) = \sum_{n=1}^{\infty} (C_n \cos{\eta_n t} + D_n \sin{\eta_n t}) \, \sin{\left ( \frac{n \pi}{L} x\right )} $$
You find the constants from the initial conditions. First, the condition that the initial string position is $f(x)$:
$$f(x) = \sum_{n=1}^{\infty} C_n \, \sin{\left ( \frac{n \pi}{L} x\right )} \implies C_n = \frac{2}{L} \int_0^L dx \, f(x) \sin{\left ( \frac{n \pi}{L} x\right )} $$
Next, the initial string velocity $u_t(x,0)=0$ implies that
$$0 = -\sum_{n=1}^{\infty} \eta_n D_n \, \sin{\left ( \frac{n \pi}{L} x\right )} \implies D_n=0$$
Thus the solution to your problem is
$$u(x,t) = \sum_{n=1}^{\infty} C_n \cos{\eta_n t} \, \sin{\left ( \frac{n \pi}{L} x\right )} $$
Best Answer
At $x=L$, at rest, the forces acting on the movable hinge (positive downwards) will be
$T$, directed along the string;
$mg/2+k(v-h)$, directed along the vertical.
Normally, the end movement $v$ is taken to be much smaller than $L$.
Therefore $T$ is considered to be constant, so that its vertical projection would be approximated by $$ T_v = T{v \over L} $$
So that at equilibrium we shall have $$ mg/2 + k(v - h) + T_v = mg/2 + k(v - h) + T{v \over L} = 0 $$ which means $$ v(L,0) = {{ - mg/2 + kh} \over {k + {T \over L}}} $$ and this answers your first question.
Now consider that a little vertical starting displacement of the end at $x=L$, if much smaller than $L$, can be neglected, i.e. that $kh=mg/2$, which is to tell that the offset positioning of the spring is just to compensate the (half) weight of the string.
In conclusion at $x=L$ you may consider to have just a reaction (on the side of the string) of $-kv$.
The second step is that an elastic clamp as the above can be "bartered" with a little increase $\Delta L$ of the length $L$ and clamp the string to a fixed hinge there.
That's because near the end the displacement $v$ is even smaller than in the central part of string where it is already taken to be much less than $L$.
That allows to put $$ kv \approx T{{dv} \over {dx}} \approx T{v \over {\Delta L}}\quad \Rightarrow \quad \Delta L \approx {T \over k} $$ if the spring is stiff enough to give $\Delta L << L$. This allows as well to neglect the effect of the dynamic reaction of the final part of the spring.
If instead the spring is quite soft then we cannot adopt the approximation above: there will be reflections at the end .
Analog of the mechanical system above is the transmission along an electrical line of finite length, terminating on a capacitive load. This goes under the name of Telegrapher's Equation and you can find for that various sources for how to solve the problem.
--- clarifications ---
As a very concise reply to your comments:
a) The modelling of the tansverse vibration of a "normal" elastic string is derived as shown in this article as (using your notation) $$ \rho v_{\,t\,t} = Tv_{\,x\,x} $$ and this is analogous to the Telegrapher's equation for a lossless line.
b) The presence of the further term $-r v$ in the equation that you propose $$ \rho v_{\,t\,t} = Tv_{\,x\,x} - rv $$ corresponds to a string that lies on an elastic "bed" (with negligible mass and shear).
So the mechanical system is in any case lossless.
c) The $v(0,t) = \sin t$ is the exciting signal fed at $x=0$ in terms of an imposed displacement $v(t)$..
That means that the actual mechanical model underlying your mathematical model is the following
that is $$ \left\{ \matrix{ T_{\,n} (x) = T\,v_{\,x} \hfill \cr T_{\,n} (x + dx) - T_{\,n} (x) - r\,v\,dx = \rho dx\,v_{\,t\,t} \quad \Rightarrow \hfill \cr \Rightarrow \quad \rho v_{\,t\,t} = Tv_{\,x\,x} - r\,v \hfill \cr v(0,t) = U(t)\sin t \hfill \cr 0 = - T_{\,n} (L) - k\,v(L) = T\,v_{\,x} (L) + k\,v(L) \hfill \cr} \right. $$ where $U(t)$ denotes the Unit Step function.
Clearly this physical model is lossless.
d) For finding the steady state solution we adopt the method described in the Telegrapher's equation article and well known to electrical engineers,
and which essentially consists in taking the Fourier transform. We put $$ v(x,t) =\Re \left( {\hat V(x)\,e^{\,j\omega t} } \right) = V(x)e^{\,j\left( {\omega t + \alpha \left( x \right)} \right)} $$ where $\hat V(x)$ is in general complex, and replace in the system above.
Now that the system is in the correct shape (if I understood properly your post) can you proceed with the math part ?