24. (15 points) Given the parabola $y^{2}=\operatorname{tar}(0<a<1)$ with focus $\vec{F}$, a semicircle is constructed above the $x$-axis with center $A(a+4,0)$ and radius $|A F|$, intersecting the parabola at two distinct points $M$ and $N$. Let $P$ be the midpoint of segment $M N$.
(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 the reason.
24. (1) From the given, we have $F(a, 0)$, and the semicircle is
$$
[x-(a+4)]^{2}+y^{2}=16(y \geqslant 0) \text {. }
$$

Substituting $y^{2}=4 a x$ into the equation, we get
$$
x^{2}-2(4-a) x+a^{2}+8 a=0 \text {. }
$$

Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$. Then, by the definition of the parabola, 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
$$
2|P F|=|M F|+|N F| \text {. }
$$

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

thus $|P F|=\left|P^{\prime} P\right|$. This indicates that point $P$ should be on the parabola.

However, it is known that $P$ is the midpoint of segment $M N$, i.e., $P$ is not on the parabola.
$\therefore$ There does not exist a value of $a$ such that $|M F|,|P F|,|N F|$ form an arithmetic sequence.