In short, row reduced echelon form(RREF) of a matrix $A$ is such that
i) Every leading entry is 1
ii) Any nonzero rows are above zero rows
iii) any leading entry is strictly to the right of any leading entries above that row
iv) any other entry in a column containing a leading entry is 0 except for the leading entry.
So it does not have to be put in augmented matrix $[A|b]$ to get a RRE form. You are comparing RRE form of matrix $A$ and $[A|b]$.
To see why the statement is true, suppose that you put the matrix $[A|b]$ into RRE form, so you have a matrix E. If E contains a leading entry in its last column, in terms of system of equations, what does it say? And what is the condition for E to not have any leading entry in last column?
Note: If RRE form of $[A|b]$ does contain a leading entry, then it is different from that of $A$. Also, note that RRE form of $[A|b]$ is m by n+1 whereas that of $A$ is m by n.
Solve:
$x+y=1$
$x+y=2$
Then we have
$\
A =
\left( {\begin{array}{cc}
1 & 1 \\
1 & 1
\end{array} } \right)
$
$\
b =
\left( {\begin{array}{cc}
1 \\
2
\end{array} } \right)
$
and $Ax=b$
If we turn A into RREF, we get
$\
E =
\left( {\begin{array}{cc}
1 & 1 \\
0 & 0
\end{array} } \right)
$
So A has rank 1
and if we put $[A|b]$ into RRE form, we get
$\
E' =
\left( {\begin{array}{cc}
1 & 1 & 0 \\
0 & 0 & 1
\end{array} } \right)
$
So augmented matrix has rank 2. Observe what last row says in terms of equations.
The pivot column in the hint can refer to a column that has a leading entry. You don't need to transform a matrix $A$ to its reduced row echelon form to see whether it has solutions. A row echelon form is enough.
Even if you transform it to its reduced row echelon form, if the last column is a pivot column, the system has no solution. You can still get all zeros for the other entries except the last one. In this case, the last equation is just $0=1$.
Here is an example for you:
$$ \left[
\begin{array}{cc|c}
1&2&3\\
1&1&0\\
1&3&1
\end{array}
\right] $$
The row echelon form of the above augmented matrix is:
$$ \left[
\begin{array}{cc|c}
1&2&3\\
0&-1&-3\\
0&0&-5
\end{array}
\right] $$
You can already see that the last column is a pivot column because it has a leading entry.
If you prefer, you can find its reduced row echelon form:
$$ \left[
\begin{array}{cc|c}
1&0&0\\
0&1&0\\
0&0&1
\end{array}
\right] $$
Best Answer
You have the augmented system $$\left[\begin{array}{cc|c}1&2&8\\1&-1&2\\1&1&4 \end{array}\right]$$ Then we get $$\left[\begin{array}{cc|c}1&2&8\\1&-1&2\\1&1&4 \end{array}\right] \leadsto \left[\begin{array}{cc|c}1&2&8\\0&-3&-6\\0&-1&-4 \end{array}\right]\leadsto\left[\begin{array}{cc|c}1&2&8\\0&1&2\\0&1&4 \end{array}\right]$$ But this is impossible to solve in two variables since from the third equation we get $y = 4$ and from the second $y = 2$.