Rewriting maximization and minimization of $|2x_1-3x_2|$ as linear programs 
$$\begin{array}{ll} \text{maximize} & |2x_1-3x_2|\\ \text{subject to} & 4x_1+x_2 \le 4\\ & 2x_1-x_2 \le 0.5\\ & x_1,x_2 \ge 0\end{array} \tag{Problem 1}$$



$$\begin{array}{ll} \text{minimize} & |2x_1-3x_2|\\ \text{subject to} & 4x_1+x_2 \le 4\\ & 2x_1-x_2 \le 0.5\\ & x_1,x_2 \ge 0\end{array} \tag{Problem 2}$$


According to the question, one of them can be rewritten as a linear program (LP), and the other one cannot. The question asks the reader to determine which one can be rewritten as an LP, and why the other one cannot.
I've never converted to an LP an optimization problem whose objective function contains an absolute value. So, I think I have no way of determining the one that can be rewritten as an LP. But, I know the simplex method and I can solve an LP using it.
 A: For problem $2$: replace the objective function by a new variable $t\ge 0$, and add the constraints
\begin{cases}
2x_1-3x_2 \le t \\
-2x_1 +3x_2 \le t
\end{cases}
These constraints are equivalent to $-t \le 2x_1-3x_2 \le t$, and so the absolute value is taken into account.
Since you are minimizing $t$, you are minimizing $\mid 2x_1-3x_2 \mid $ indeed.
It is not difficult to see that this does not work if you are maximizing $t$.
A: Let $f (x, y) := | 2 x - 3 y |$. Note that $f$ is convex. Its epigraph is the convex set
$$\left\{ (x,y,z) \in \mathbb R^3 \mid z \geq 2 x - 3 y \land z \geq - 2 x + 3 y \right\}$$
Hence, the minimization problem ("problem 2") can be written as the following linear program
$$\boxed{\begin{array}{ll} \text{minimize} & z\\ \text{subject to} & \begin{bmatrix} 2 & -3 & -1\\ -2 & 3 & -1\\ 4 & 1 & 0\\ 4 & -2 & 0\end{bmatrix} \begin{bmatrix} x\\ y\\ z\\ \end{bmatrix} \leq \begin{bmatrix} 0\\ 0\\ 4\\ 1\end{bmatrix}\\ & x, y, z \geq 0\end{array}}$$
The maximization problem ("problem 1") can be written in minimization form as follows
$$\begin{array}{ll} \text{minimize} & -|2 x - 3 y|\\ \text{subject to} & 4x + y \leq 4\\ & 4x - 2y \leq 1\\ & x, y \geq 0\end{array}$$
where the objective function $g (x, y) := - f (x, y)$ is concave. Thus, the epigraph of $g$ is non-convex and cannot be defined by the intersection of half-spaces. Although we cannot write "problem 1" as a linear program, we can write is as two linear programs, namely,
$$\begin{array}{ll} \text{minimize} & -2 x + 3 y\\ \text{subject to} & 4x + y \leq 4\\ & 4x - 2y \leq 1\\ & 2 x - 3 y \geq 0\\ & x, y \ge 0\end{array}$$
and
$$\begin{array}{ll} \text{minimize} & 2 x - 3 y\\ \text{subject to} & 4x + y \leq 4\\ & 4x - 2y \leq 1\\ & 2 x - 3 y \leq 0\\ & x, y \ge 0\end{array}$$
We solve both linear programs and then take the minimum of the two minima. Reversing the sign, we obtain the maximum of "problem 1".

Addendum
From Boyd & Vandenberghe:


