13. 6:10 ** Given the parabola $y^{2}=4 a x(0<a<1)$ with focus $F$, a circle is drawn above the $x$-axis with center $A(a+4,0)$ and radius $|A F|$, intersecting the parabola at two different points $M$ and $N$. Let $P$ be the midpoint of $M N$.
(1) Find the value of $|M F|+|N F|$;
(2) Does there exist a value of $a$ such that $|M F|, |P F|, |N F|$ form an arithmetic sequence? If so, find the value of $a$; if not, explain why.
(1) As shown in the figure, from the given information, the focus is $F(a, 0)$, and the equation of the semicircle is $[x-(a+4)]^{2}+y^{2}=16(y \geqslant 0)$. Substituting $y^{2}=1 a x$ into the semicircle equation yields $x^{2}-2(4-a) x+a^{2}+8 a=0$. Let the coordinates of $M$ and $N$ be $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$, respectively. Then, by the definition of the parabola and Vieta's formulas, we have
$$
\begin{aligned}
|M F|+|N F| & =x_{1}+x_{2}+2 a \\
& =2(4-a)+2 a=8 .
\end{aligned}
$$
(2) If $|M F|, |P F|, |N F|$ form an arithmetic sequence, then $2|P F|=|M F|+|N F|$.

On the other hand, let the projections of $M, P, N$ on the directrix of the parabola be $M^{\prime}, P^{\prime}, N^{\prime}$, respectively. Then, in the right trapezoid $M^{\prime} M N N^{\prime}$, $P^{\prime} P$ is the midline, so we have
$$
2\left|P^{\prime} P\right|=\left|M^{\prime} M\right|+\left|N^{\prime} N\right|=|F M|+|F N|,
$$

thus $|P F|=\left|P^{\prime} P\right|$, which indicates that $P$ lies on the parabola. However, it is known that $P$ is the midpoint of segment $M N$, meaning $P$ does not lie on the parabola. Therefore, there is no value of $a$ that makes $|M F|, |P F|, |N F|$ form an arithmetic sequence.