$$u_{tt} - u_{xx} = 0$$
The general solution (without boundary condition) is :
$$u(x,y)=f(x+t)+g(x+t)$$
The functions $f$ and $g$ are not necessarily the same.
The condition $u(x,0) = \phi(x)=f(x)+g(x)$ implies
$\quad\begin{cases}
f(x)=\frac12\phi(x)+h(x) \\
g(x)=\frac12\phi(x)-h(x)
\end{cases}$
$$ u(x,t)=\frac12\phi(x+t)+h(x+t)+\frac12\phi(x-t)-h(x-t)$$
$h(x)$ is an arbitrary function to be determined by the condition $u_t(x,0)=0$
$$ u_t(x,t)=\frac12\phi'(x+t)+h'(x+t)-\frac12\phi'(x-t)+h'(x-t)$$
$$ u_t(x,0)=\frac12\phi'(x)+h'(x)-\frac12\phi'(x)+h'(x)$$
$$u_t(x,0)=0=2h'(x) \quad\implies\quad h(x)=C$$
Your solution is confirmed :
$$ u(x,t)=\frac12\phi(x+t)+\frac12\phi(x-t)$$
The function $\phi$ is a given piecewise function :
$$\phi(x) = \begin{cases}
1 & |x| \leq 1 \\
0 & |x| > 1 \\
\end{cases}$$
$$ u(x,t)=\frac12\begin{cases}
1 & x+t \leq 1 \\
1 & -x-t \leq 1 \\
0 & x+t > 1 \\
0 & -x-t > 1 \\
\end{cases}+
\frac12\begin{cases}
1 & x-t \leq 1 \\
1 & -x+t \leq 1 \\
0 & x-t > 1 \\
0 & -x+t > 1 \\
\end{cases}$$
$$ u(x,t)=\frac12\begin{cases}
1 & x \leq 1-t \\
1 & x \geq -1-t \\
0 & x > 1-t \\
0 & x < -1-t \\
\end{cases}+
\frac12\begin{cases}
1 & x \leq 1+t \\
1 & x \geq -1+t \\
0 & x > 1+t \\
0 & x < -1+t \\
\end{cases}$$
We have to consider several regions limited by
$x=1-t \quad;\quad x=1+t \quad;\quad x=-1-t \quad;\quad x=-1+t\quad$
For $\quad \boxed{0<t<1}$ :
Case $\quad x>1+t \:: \quad u=\frac12(0)+\frac12(0)=0$.
Case $\quad 1-t<x\leq 1+t \: : \quad u=\frac12(0)+\frac12(1)=\frac12$.
Case $\quad -1+t<x\leq 1-t \: : \quad u=\frac12(1)+\frac12(1)=1$.
Case $\quad -1-t\leq x< -1+t \: : \quad u=\frac12(1)+\frac12(0)=\frac12$.
Case $\quad x\leq -1-t \: : \quad u=\frac12(0)+\frac12(0)=0$.
$$u(x,t)=\begin{cases}
0 && x>1+t \\
\frac12 && 1-t<x\leq 1+t \\
1 && -1+t<x\leq 1-t\\
\frac12 && -1-t\leq x< -1+t\\
0 && x\leq -1-t
\end{cases}$$
The above formulas are not valid for $t>1$ which is outside the domain of study specified in the wording of the question.
IN ADDITION :
For $\quad \boxed{t>1}$ :
$$u(x,t)=\begin{cases}
0 && x>1+t \\
\frac12 && -1+t<x\leq 1+t \\
0 && 1-t<x\leq -1+t\\
\frac12 && -1-t\leq x< 1-t\\
0 && x\leq -1-t
\end{cases}$$
You should only ask questions about a single problem. This is not the place for a homework dump.
With that being said, I'm going to solve the first problem. First, define the primitive of $u_t(x,0)=g(x)$ as
$$ G(s) = \int g(s)\ ds = \begin{cases} \dfrac{1}{\pi}\sin\big(\pi(s-1)\big), & 1 < s < 2 \\ 0, & \text{otherwise} \end{cases} $$
Then the solution given by d'Alembert's formula is
$$ u(x,t) = \frac{G(x+t) - G(x-t)}{2} $$
where you have to check case by case for both $x+t$ and $x-t$. Below is a graph ($x$ vs. $t$) of the 4 characteristic lines going through $x=1$ and $x=2$.
The numbered regions are as follows:
\begin{array}{rrr}
\text{I}: && x - t < 1, && x + t < 1 \\
\text{II}: && x - t < 1, && 1 < x + t < 2 \\
\text{III}: && x - 1 < 1, && x + t > 2 \\
\text{IV}: && 1 < x - t < 2, && 1 < x + t < 2 \\
\text{V}: && 1 < x - t < 2, && x + t > 2 \\
\text{VI}: && x - t > 2, && x + t > 2 \\
\end{array}
Then, you can simplify the solution
$$ u(x,t) = \begin{cases}
0, && (x,t) \in \text{I, III, VI} \\
\dfrac{1}{2\pi}\sin\big(\pi(x+t-1)\big), && (x,t) \in \text{II} \\
\dfrac{1}{2\pi}\Big[\sin\big(\pi(x+t-1)\big) - \sin\big(\pi(x-t-1)\big) \Big], && (x,t) \in \text{IV} \\
-\dfrac{1}{2\pi}\sin\big(\pi(x-t-1)\big), && (x,t) \in \text{V}
\end{cases} $$
Best Answer
$f$ is continuous, and $g$ has discontinuity of the first kind in zero, so we can use d'Alembert's formula: $u(t,x) = \frac{1}{2} \int_{x-t}^{x+t} g(z) dz = \frac{1}{2} (\int_{x-t}^{0} g(z)dz+\int_{0}^{x+t} g(z)dz)$. In order to calculate this integral we consider three sets: $$K = \{(t,x)|t>0,x-t<0,x+t<0\}\\ M = \{(t,x)|t>0,x-t<0,x+t>0\}\\ P = \{(t,x)|t>0,x-t>0,x+t>0\}$$
And we obtain $u(t,x)=\left\{\begin{matrix} 0, & (t,x) \in K, \\ \frac{x+t}{2}, & (t,x) \in M, \\ t, & (t,x) \in N, \end{matrix}\right.$
It is easy to prove that $u$ belongs to the class $C([0,\infty)\times\mathbb{R})$. But this functions is not continuously differentiable.
Also, we can prove that this problem has unique solution from class $C([0,\infty)\times\mathbb{R})$.