1. Lemma Proof:
We start by proving the lemma that states: In triangle $ABC$ with incenter $I$, let $BI$ intersect $AC$ at $E$ and $CI$ intersect $AB$ at $F$. Let $M$ be the midpoint of $BC$. Then the $A$-symmedian, $EF$, and $IM$ are concurrent.

To prove this, we use barycentric coordinates. Let the coordinates of $A$, $B$, and $C$ be $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ respectively. The incenter $I$ has coordinates $(a:b:c)$. The coordinates of $E$ and $F$ can be determined as follows:
- $E$ lies on $BI$ and $AC$, so its coordinates are $(0:b:c)$.
- $F$ lies on $CI$ and $AB$, so its coordinates are $(a:0:c)$.

The midpoint $M$ of $BC$ has coordinates $(0:1:1)$.

Now, let $X$ be the point of intersection of the $A$-symmedian and $IM$. Assume $X$ has coordinates $(x:y:z)$. Since $X$ lies on the $A$-symmedian, we have $y:z = b^2:c^2$, implying $z = \frac{c^2y}{b^2}$.

Since $X$ also lies on $IM$, we have:
$$0 = \begin{vmatrix} x & y & z \\ a & b & c \\ 0 & 1 & 1 \end{vmatrix} = x(b-c) - ya + \frac{ac^2y}{b^2}$$
Simplifying, we get:
$$x(b-c) = y \left(a - \frac{ac^2}{b^2}\right) \implies x:y = \frac{a(b^2-c^2)}{b^2(b-c)} = \frac{a(b+c)}{b^2}$$
Thus, the coordinates of $X$ are $(ab+ac:b^2:c^2)$.

To prove $X$, $E$, and $F$ are collinear, we need the determinant to be zero:
$$\begin{vmatrix} a & 0 & c \\ a & b & 0 \\ ab+ac & b^2 & c^2 \end{vmatrix} = abc^2 + c(ab^2 - ab^2 - abc) = 0$$
Hence, $X$, $E$, and $F$ are collinear.

2. Main Problem:
Given the lemma, we know that $AQ$ is the $A$-symmedian of $\Delta ABC$. Let $AP$ and $AQ$ intersect $\odot(ABC)$ at $T$ and $V$ respectively, and let $EF$ intersect $\odot(ABC)$ at $X$ and $Y$. Let $VT$ intersect $BC$ at $K$.

Since $-1 = (A, V; B, C) \overset{T}{=} (\overline{AT} \cap \overline{BC}, K; B, C)$, it follows that $K \in \overline{XY}$. Thus, $(B, C)$, $(X, Y)$, and $(T, V)$ are pairs of an involution $\Phi$. Projecting $\Phi$ from $A$ onto $\overline{EF}$, we get that $(X, Y)$, $(E, F)$, and $(P, Q)$ are pairs of an involution. Hence, $\odot(ABC)$, $\odot(AEF)$, and $\odot(APQ)$ are coaxial.

Therefore, $A$, $P$, $Q$, and $R$ are concyclic. $\blacksquare$