We know that the general solution of the wave equation $$\frac{\partial ^2u}{\partial t^2}=c^2\frac{\partial ^2 u}{\partial x^2}$$ is $$u(x,t)=f(x-ct)+g(x+ct),$$ i.e., a sum of a wave travelling left, and a sum of a wave travelling right. In the problem, we have $c=1,$ so the general solution would be of the form $$u(x,t)=f(x-t)+g(x+t).$$D'Alembert's formula tells us that, based on the initial conditions, we have
\begin{align*}
f(\xi)&=\xi+\frac12,\quad\xi\ge 0\\
g(\zeta)&=\zeta+\frac12,\quad\zeta\ge 0.
\end{align*}
Thus, we have
\begin{align*}
f(x-t)&=x-t+\frac12,\quad x-t\ge 0\\
g(x+t)&=x+t+\frac12,\quad x+t\ge 0,
\end{align*}
yielding $$u(x,t)=2x+1,$$ on the region $x\ge t.$ Note that this doesn't have to satisfy the boundary condition, because it is imposed on the line $x=0,$ which doesn't lie in that region.
To obtain the solution on the other region, we need to extend the domain of $f$ so that it can receive negative inputs. To do so, we will use the boundary condition. Writing it out in terms of $f$ and $g,$ we have $$f'(-t)+f(-t)+g'(t)+g(t)=0,\quad t>0,$$or $$f'(t)+f(t)+g'(-t)+g(-t)=0,\quad t<0.$$This is now a first order ODE for $f,$ which we solve as follows:
\begin{align*}
e^tf'(t)+e^tf(t)&=-e^t\left(1-t+\frac12\right)\\
(e^tf(t))'&=-e^t\left(\frac32-t\right)\\
e^\xi f(\xi)-e^0f(0)&=-\int_0^\xi e^t\left(\frac32-t\right)dt\\
e^\xi f(\xi)&=-\left(e^\xi(2.5-\xi)-2.5\right)+\frac12\\
f(\xi)&=-(2.5-\xi)+2.5e^{-\xi}+0.5e^{-\xi}\\
f(\xi)&=\xi-2.5+3e^{-\xi},\quad \xi<0.
\end{align*}
Thus, the solution on the region $x<t$ then becomes $$u(x,t)=x-t-2.5+3e^{t-x}+x+t+0.5,$$ or $$u(x,t)=2x-2+3e^{t-x}.$$This doesn't satisfy the initial conditions, but it doesn't have to -- note that they're imposed on the line $t=0,$ which doesn't lie in the region $t>x.$
As a sanity check, plugging this into the BC, we get $$2-3e^{t}-2+3e^t=0.$$ We can also check that it satisfies the wave equation itself: $$\frac{\partial ^2u}{\partial x^2}=3e^{t-x}=\frac{\partial ^2u}{\partial t^2}.$$Thus, the solution is $$u(x,t)=\begin{cases}2x+1,&x\ge t\\2x-2+3e^{t-x},&x<t.\end{cases}$$
