Four. (25 points) Given three linear functions
$$
y=x+1, y=1-x, y=\frac{1}{2} x+b \text {. }
$$
(1) If the graphs of these three functions can form a triangle, find the range of the real number $b$;
(2) If the area of the triangle formed by the graphs of these three functions is $\frac{4}{3}$, find the value of the real number $b$.
(1) Obviously, the line $l_{1}: y=x+1$ intersects with $l_{2}: y=1-x$ at point $P(0,1)$.

Only when the line $l_{3}: y=\frac{1}{2} x+b$ passes through point $P(0,1)$, the three lines cannot form a triangle, at this time, $b=1$.
Therefore, when $b \neq 1$, the three lines can form a triangle.
Thus, the intersection point of line $l_{1}$ and $l_{3}$ is $A(2 b-2,2 b-1)$.
Solving the system $\left\{\begin{array}{l}y=1-x, \\ y=\frac{1}{2} x+b\end{array} \Rightarrow\left\{\begin{array}{l}x_{2}=-\frac{2}{3} b+\frac{2}{3}, \\ y_{2}=\frac{2}{3} b+\frac{1}{3} .\end{array}\right.\right.$
Thus, the intersection point of line $l_{2}$ and $l_{3}$ is
$$
B\left(-\frac{2}{3} b+\frac{2}{3}, \frac{2}{3} b+\frac{1}{3}\right) \text {. }
$$

Assume the three lines form $\triangle A B P$, then its area is
$$
\begin{array}{l}
S_{\triangle A B P}=\frac{1}{2}\left|x_{1}-x_{2}\right||1-b| \\
=\frac{1}{2}\left|\left(-\frac{2}{3} b+\frac{2}{3}\right)-(2 b-2)\right||1-b| \\
=\frac{4}{3}(1-b)^{2}=\frac{4}{3} .
\end{array}
$$

Thus $(1-b)^{2}=1$
$$
\Rightarrow 1-b= \pm 1 \text {. }
$$

Solving gives $b=0$ or 2.
Therefore, the real number $b=0$ or 2.