3. For the parabola $y^{2}=4 a x(a>0)$, the focus is $A$. With $B(a+4,0)$ as the center and $|A B|$ as the radius, a semicircle is drawn above the $x$-axis, intersecting the parabola at two different points $M, N$. Let $P$ be the midpoint of the segment $M N$. (I) Find the value of $|A M|+|A N|$; (II) Does there exist such an $a$ that $|A M|,|A P|,|A N|$ form an arithmetic sequence?
3. ( I) The equation of the semicircle is $[x-(a+4)]^{2}+y^{2}=16(y \geqslant 0)$, and when combined with $y^{2}=4 a x(a>0)$, eliminating $y$ yields $x^{2}-(8-$ $2 a) x+a(a+8)=0$.

Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$, then $|A M|=x_{1}+a,|A N|=x_{2}+a$, hence $|A M|+|A N|=x_{1}+x_{2}+2 a=(8-$ $2 a)+2 a=8$.
(II) If there exists $a$ such that $|A M|,|A P|,|A N|$ form an arithmetic sequence, draw perpendicular segments $M M_{1}$, $P P_{1}, N N_{1}$ from $M, P, N$ to the directrix $x=-a$, then by the definition of the parabola, we have $2|A P|=|A M|+|A N|=\left|M M_{1}\right|+\left|N N_{1}\right|=2\left|P P_{1}\right|$. Draw a line $P T / / y$-axis intersecting the parabola at $T$, and draw $T T_{1} \perp$ the directrix $x=-a$ at $T_{1}$, then $\left|P P_{1}\right|=\left|T T_{1}\right|$, thus $2|A P|=2\left|T T_{1}\right|$. Since $\left|T T_{1}\right|=|A T|$, we have $2|A P|=2|A T|$, i.e., $|A P|=|A T|$, which clearly contradicts the fact that point $P$ is not on the parabola. Therefore, there does not exist such $a$ that makes $2|A P|$ equal to $|A M|+|A N|$.