13. Given the quadratic function $f(x)=x^{2}-16 x+p+3$.
(1) If the function has a zero in the interval $[-1,1]$, find the range of real number $p$;
(2) Does there exist a constant $q(q \geq 0)$, such that when $x \in[q, 10]$, the range of $f(x)$ is an interval $D$, and the length of $D$ is $12-q$. Note: The length of the interval $[a, b](a<b)$ is $b-a$.
Solution (1) Since the axis of symmetry of the quadratic function $f(x)=x^{2}-16 x+p+3$ is $x=8$, the function $f(x)$ is monotonically decreasing in the interval $[-1,1]$. Therefore, for the function $f(x)$ to have a zero point in the interval $[-1,1]$, it must satisfy $f(-1) \cdot f(1) \leq 0$, i.e., $(1+16+p+3)(1-16+p+3) \leq 0$, solving this gives $-20 \leq p \leq 12$.
5 points
(2) Assume there exists a constant $q(q \geq 0)$ that meets the condition, and solve in three cases:
When $\left\{\begin{array}{l}q<8 \\ 8-q \geq 10-8 \\ q \geq 0\end{array}\right.$, i.e., $0 \leq q \leq 6$.

When $x=8$, the minimum value $f(8)$ is obtained; when $x=q$, the maximum value $f(q)$ is obtained, so the range of $f(x)$ is: $[f(8), f(q)]$, i.e., $\left[p-61, q^{2}-16 q+p+3\right]$, so the interval length is $q^{2}-16 q+p+3-(p-61)=q^{2}-16 q+64=12-q$.
So $q^{2}-15 q+52=0$, so $q=\frac{15 \pm \sqrt{17}}{2}$, upon inspection, $q=\frac{15+\sqrt{17}}{2}$ does not meet the condition, discard it, hence $q=\frac{15-\sqrt{17}}{2}$;
(2) When $\left\{\begin{array}{l}q<8 \\ 8-q<10-8 \\ q \geq 0\end{array}\right.$, i.e., $6 \leq q<8$.
When $x=8$, the minimum value $f(8)$ is obtained; when $x=10$, the maximum value $f(10)$ is obtained, so the range of $f(x)$ is $[f(8), f(10)]$, i.e., $[p-61, p-57]$, so the interval length is $p-57-(p-61)=4=12-q$, so $q=8$, upon inspection, it does not meet the condition, discard it;
(3) When $q \geq 8$, the function $f(x)$ is monotonically increasing in $[q, 10]$, so the range of $f(x)$ is: $[f(q), f(10)]$, i.e., $\left[q^{2}-16 q+p+3, p-57\right]$, so the interval length is $p-57-\left(q^{2}-16 q+p+3\right)=-q^{2}-16 q-60=12-q$, so $q^{2}-17 q+72=0$, so $q=8$ or $q=9$, upon inspection, $q=8$ or $q=9$ meet the condition.

In summary, there exist constants $q=8$ or $q=9, q=\frac{15-\sqrt{17}}{2}$, such that when $x \in[q, 10]$, the range of $f(x)$ is the interval $D$, and the length of $D$ is $12-q$.
20 points