Example $8 A, B, C, D, E$ five people participated in an exam, which has seven questions, all of which are true or false questions. The scoring criteria are: for each question, 1 point is awarded for a correct answer, 1 point is deducted for a wrong answer, and no points are awarded or deducted for unanswered questions. Table 1 records the answers given by $A, B, C, D, E$ five individuals. It is now known that $A, B, C, D$ each scored 2 points. Question: How many points should $E$ get? What is the correct answer for each question?
Let $x_{k}=\left\{\begin{array}{ll}1, & \text { if the conclusion of the } k \text {th question is correct; } \\ -1, & \text { if the conclusion of the } k \text {th question is incorrect, }\end{array}\right.$ where $k=1,2, \cdots, 7$.

At this point, if the conclusion is judged to be correct (i.e., marked with a “ $\checkmark$ ”), then $x_{k}$ points are obtained; if it is judged to be incorrect (i.e., marked with a “ $\times$ ”), then $-x_{k}$ points are obtained.
Since $A, B, C, D$ each scored 2 points, we can derive the system of equations:
$$
\left\{\begin{array}{l}
x_{1}+0 \cdot x_{2}-x_{3}+x_{4}-x_{5}+x_{6}+x_{7}=2, \\
x_{1}-x_{2}+x_{3}+x_{4}-x_{5}-x_{6}+0 \cdot x_{7}=2, \\
0 \cdot x_{1}+x_{2}-x_{3}-x_{4}+x_{5}-x_{6}+x_{7}=2, \\
-x_{1}-x_{2}-x_{3}+x_{4}+x_{5}+0 \cdot x_{6}-x_{7}=2 .
\end{array}\right.
$$

Adding these four equations yields
$$
x_{1}-x_{2}-2 x_{3}+2 x_{4}-x_{6}+x_{7}=8 \text {. }
$$

Noting that $x_{i}= \pm 1(i=1,2, \cdots, 7)$.
Therefore, the left side of the above equation is less than or equal to 8, while the right side is 8, hence
$$
\begin{array}{l}
x_{1}=1, x_{2}=-1, x_{3}=-1, \\
x_{4}=1, x_{6}=-1, x_{7}=1 .
\end{array}
$$

Substituting these results into the first equation of the system, we get $x_{5}=1$.
Thus, questions 1, 4, 5, and 7 are correct, while questions 2, 3, and 6 are incorrect.
Therefore, according to the problem, $E$ scores 4 points.