If the row echelon form of a linear equation system has a line with zero entries, then it has more than a solution. 
If the row echelon form of a linear equation system has a line with zero entries, then it has
  more than one solution.

Let
\begin{align}
\text{I}& \qquad 3x_1&=2\\
\text{II}& \qquad 0x_1&=0
\end{align} a linear equation system in row echelon form and a line with zero entries.
If we solve this equation system, the solution $\mathcal{L}$ would be $\mathcal{L}=\{x_1=\frac23\}$. Contradiction. The statement is wrong.

This looks way too easy. Is there anything I might forgot about?
EDIT:

If the row echelon form of a linear system of equations has a column with zero entries, it has no solution.

This statement is wrong (too), right?
 A: Let's start from the linear system $Ax=b$. You can form the matrix $[A\mid b]$ and perform row reduction to get, say, the row echelon form $[U\mid c]$.
The system has at least a solution if and only if $c$ is not a pivot column. In this case, 

the system has a single solution if and only if every column in $U$ is a pivot column

independently on the number of zero rows in $[U\mid c]$.
If the system has more unknowns than equations, then, as soon as it has a solution it will have infinitely many of them, because it's impossible that every column in $U$ is a pivot column.
The case when the system has a single solution and there are zero rows in $[U|c]$ is possible when the system has more equations than unknowns. The example you present is indeed one of these.
The statement “if the row echelon form has a zero column then the system has no solution” is clearly false. If the system has a solution at all, it will have infinitely many of them, because the unknown corresponding to the zero column can get whatever value you want.
