13. $A, B, C, D, E$ five people participate in an exam, which has 7 questions, all of which are true/false questions. The scoring rule is: for each question, a correct answer earns 1 point, a wrong answer deducts 1 point, and no answer neither earns nor deducts points. Figure 15-12 records the answers of $A, B, C, D, E$ five people. It is known that $A, B, C, D$ each scored 2 points. What should $E$ score? What is the answer to each question?
13. Let $x_{k}=\left\{\begin{array}{l}1, \text { if the } k \text {th problem is correct, } \\ -1, \text { if the } k \text {th problem is incorrect, }\end{array}\right.$ $k=1,2, \cdots, 7$. Thus, when the $k$th problem is correct, i.e., $x_{k}=1$. If it is judged as correct (i.e., marked with a “ $\checkmark$ ”), then $x_{k}$ points are scored; if it is judged as incorrect (i.e., marked with a “ $\times$ ”), then $-x_{k}$ points are scored. When the $k$th problem is incorrect, i.e., $x_{k}=-1$, if it is judged as correct, then $x_{k}$ points are scored; if it is judged as incorrect, then $-x_{k}$ points are scored. Since $A, B, C, D$ each scored 2 points, we can derive the system of equations:
$$
\left\{\begin{array}{l}
x_{1}+0 x_{2}-x_{3}+x_{1}-x_{5}+x_{6}+x_{7}=2, \\
x_{1}-x_{2}+x_{3}+x_{4}-x_{5}-x_{6}+0 x_{7}=2, \\
0 x_{1}+x_{2}-x_{3}-x_{4}+x_{5}-x_{6}+x_{7}=2, \\
-x_{1}-x_{2}-x_{3}+x_{4}+x_{5}+0 x_{6}-x_{7}=2 .
\end{array}\right.
$$

Adding these four equations, we get
$$
x_{1}-x_{2}-2 x_{3}+2 x_{4}+0 x_{5}-x_{6}+x_{7}=8 \text {. }
$$

Noting that $x_{i} \pm(i=1,2, \cdots, 7)$, the left side of the above equation is $\leqslant 8$, and the right side is $=8$, hence $x_{1}=1, x_{2}=-1, x_{3}=-1, x_{4}=$ $1, x_{6}=-1, x_{7}=1$. Substituting these results into the first equation of the system, we get $x_{5}=1$, so among the 7 problems, the 1st, 4th, 5th, and 7th problems are correct, and the 2nd, 3rd, and 6th problems are incorrect. According to the problem, $E$ scored 4 points.