The general solution for $u$ is $ u = f(x_0) = f(x-ut)$, now consider the first set of initial data.

When t = 0,

Case1. If $x < -a$, then we have $u = 1$

Case2. If $x > a$, then we have $u = -1$

Case3. If $-a\le x \le a$, then we have $u =- \frac{x}{a}$.

\begin{equation}

u = -\frac{x-ut}{a}\rightarrow u = \frac{x}{t-a}

\end{equation}

Then the solution looks like

\begin{align}

u(x,t)=&\left\{\begin{aligned}

1& && x<-a+t,\\

\frac{x}{t-a}& && -a+t\le x \le a-t,\\

-1& && x>a-t;

\end{aligned}\right.

\\

\end{align}

Thus as t increase, -a+t and a-t runs toward each other, the the solution graph is squeezing. The solution is define for $0<t<a$.