Four. (25 points) Let the linear function $y=k x+b(k \neq 0)$ intersect the positive half-axes of the $x$-axis and $y$-axis at points $E$ and $F$, respectively, and intersect the linear function $y=2 x-1$ at point $A(m, 2)$, with $A$ being the midpoint of $E F$.
(1) Find the expression of the linear function;
(2) If the graph of the linear function $y=2 x-1$ intersects the $x$-axis at point $P$, find the area of $\triangle A P E$.

保留了原文的换行和格式。
(1) Since point $A(m, 2)$ lies on the graph of $y=2 x-1$, we have,
$$
2=2 m-1 \Rightarrow m=\frac{3}{2} \Rightarrow A\left(\frac{3}{2}, 2\right) \text {. }
$$

Since $A$ is the midpoint of $E F$, the points are $E(3,0), F(0,4)$.
Thus, the required linear function expression is
$$
y=-\frac{4}{3} x+4 \text {. }
$$
(2) According to the problem, point $P\left(\frac{1}{2}, 0\right)$. Therefore,
$$
S_{\triangle P E}=\frac{1}{2}\left(3-\frac{1}{2}\right) \times 2=\frac{5}{2} .
$$