Three. (25 points) Given the line $y=-\frac{3}{4} x+3$ intersects the $x$-axis at point $C$ and the $y$-axis at point $B$. The parabola $y=a x^{2}+ \frac{3}{4} x+c$ passes through points $B$ and $C$, and intersects the $x$-axis at a second point $A$.
(1) As shown in Figure 4, $E$ is a moving point on the parabola above and to the right of line $BC$. When the area of $\triangle B E C$ is maximized, find the equation of the tangent line $l$ at point $E$ and the maximum area of $\triangle B E C$.
(2) Under the conclusion of (1), draw a line parallel to the $y$-axis through point $E$, intersecting line $BC$ at point $M$. Connect $AM$, and let $Q$ be a moving point on the axis of symmetry of the parabola. Does there exist a point $P$ on the parabola such that the quadrilateral $P Q A M$ is a parallelogram? If so, directly write the coordinates of point $P$; if not, explain the reason.
(3) As shown in Figure 5, point $N(1,0)$. Rotate segment $ON$ counterclockwise around point $O$ to get $ON'$, with the rotation angle $\alpha\left(0^{\circ}<\alpha<90^{\circ}\right)$. Find the minimum value of $N' C + \frac{1}{3} N' B$.
(1) From the problem, we know
$$
A(-2,0), B(0,3), C(4,0) \text {. }
$$

Thus, the equation of the parabola is
$$
y=-\frac{3}{8} x^{2}+\frac{3}{4} x+3 \text {. }
$$

Draw a line parallel to the $y$-axis through point $E$, intersecting $B C$ at point $H$.
Let $E\left(x,-\frac{3}{8} x^{2}+\frac{3}{4} x+3\right)$.
Then $H\left(x,-\frac{3}{4} x+3\right)$.
Therefore, $S_{\triangle B E C}=\frac{1}{2} E H \cdot O C$
$=\frac{1}{2} \times 4\left(-\frac{3}{8} x^{2}+\frac{3}{4} x+3+\frac{3}{4} x-3\right)$
$=-\frac{3}{4} x^{2}+3 x$.
Since $-\frac{3}{4}<0$, the maximum value of $S_{\triangle B E C}$ is 3, at which point, $E(2,3)$.
Let the line $l: y=k x+b$.
Since $l / / B C$, we have $k=-\frac{3}{4}$.
From point $E(2,3)$ lying on line $l$, we get $b=\frac{9}{2}$.
Therefore, the line $l: y=-\frac{3}{4} x+\frac{9}{2}$.
(2) It exists.
It is easy to know $M\left(2, \frac{3}{2}\right)$, and the axis of symmetry of the parabola is the line $x=1$.
Let point $P(m, n), Q(1, s)$. Then
$$
n=-\frac{3}{8} m^{2}+\frac{3}{4} m+3 \text {. }
$$
(i) If $A M$ is a side of the parallelogram.

Notice that point $A$ is first moved 4 units to the right, then  $\frac{3}{2}$ units up to get point $M$.

Thus, point $Q(P)$ should first be moved 4 units to the right, then $\frac{3}{2}$ units up to get $P(Q)$.
Therefore, $m=1+4=5$ or $m=1-4=-3$.
Hence, $P\left(5,-\frac{21}{8}\right)$ or $\left(-3,-\frac{21}{8}\right)$.
(ii) If $A M$ is a diagonal of the parallelogram.
By the midpoint formula, we get
$$
-2+2=m+1 \Rightarrow m=-1 \text {. }
$$

Thus, $P\left(-1, \frac{15}{8}\right)$.
In summary, the coordinates of point $P$ are
$$
\left(5,-\frac{21}{8}\right) \text { or }\left(-3,-\frac{21}{8}\right) \text { or }\left(-1, \frac{15}{8}\right) \text {. }
$$
(3) As shown in Figure 10, take a point $M^{\prime}$ on the $y$-axis such that $O M^{\prime}=\frac{1}{3}$, and connect $C M^{\prime}$;
Take a point $N^{\prime}$ on $C M^{\prime}$ such that $O N^{\prime}=O N$.
Since $O N^{\prime}=1$,
$$
O M^{\prime} \cdot O B=\frac{1}{3} \times 3
$$
$$
=1 \text {, }
$$

we know
$$
\begin{array}{l}
O N^{\prime 2}=O M^{\prime} \cdot O B \\
\Rightarrow \frac{O N^{\prime}}{O M^{\prime}}=\frac{O B}{O N^{\prime}} . \\
\text { Also, } \angle B O N^{\prime}=\angle M^{\prime} O N^{\prime}, \text { thus, } \\
\triangle M^{\prime} O N^{\prime} \backsim \triangle N^{\prime} O B \\
\Rightarrow \frac{M^{\prime} N^{\prime}}{B N^{\prime}}=\frac{O N^{\prime}}{O B}=\frac{1}{3} \\
\Rightarrow M^{\prime} N^{\prime}=\frac{1}{3} B N^{\prime} \\
\Rightarrow C N^{\prime}+\frac{1}{3} B N^{\prime}=C N^{\prime}+N^{\prime} M^{\prime}=C M^{\prime} .
\end{array}
$$

At this point, $C N^{\prime}+\frac{1}{3} B N^{\prime}$ is minimized, which is when $C, M^{\prime}, N^{\prime}$ are collinear, due to the shortest distance between two points being a straight line.
Thus, the minimum value sought is
$$
C M^{\prime}=\sqrt{4^{2}+\left(\frac{1}{3}\right)^{2}}=\frac{\sqrt{145}}{3} .
$$