Proof 1. (1) Let $A_1$ be the reflection of $A$ across $B$. By the given conditions, $KPA_1A$ forms an isosceles trapezoid, thus $\angle APB = \angle A_1KB$.
Similarly, let $A_2$ be the reflection of $A$ across $C$. We have $\angle AQC = \angle A_2KC = \angle A_2KB$.
Hence, the desired conclusion is equivalent to $\angle BTC = \angle A_1KA_2$.
Perform a homothety centered at $A$ with a ratio of $1/2$ on $\triangle A_1KA_2$, mapping $A_1$ to $B$, $A_2$ to $C$, and $K$ to the midpoint of $AK$ (denoted as $M$). The proposition then transforms to $\angle BTC = \angle BMC$, i.e., points $B, C, M$, and $T$ are concyclic.
By the Power of a Point Theorem, this is equivalent to proving $KB \cdot KC = KM \cdot KT$. If we take $X$ as the midpoint of $KT$, it is equivalent to proving $KB \cdot KC = KA \cdot KX$, i.e., points $A, B, C$, and $X$ are concyclic.
We note that the perpendicular bisector of segment $KP$ passes through point $B$ and is perpendicular to $AB$, and the perpendicular bisector of $KQ$ passes through $C$ and is perpendicular to $AC$. Hence, the circumcenter of $\triangle PQK$ (denoted as $O$) is the antipodal point of $A$ on the unit circle. From $OX \perp KT$, we have $\angle OXA = \frac{\pi}{2}$, and thus $X$ lies on the circle with diameter $AO$ (the circumcircle of $\triangle ABC$). Proof completed.

(2) We have
$$\begin{aligned} \frac{BP}{CQ} &= \frac{BK}{CK} = \frac{S_{\triangle BTK}}{S_{\triangle CTK}} = \frac{BT}{CT} \cdot \frac{\sin \angle BTK}{\sin \angle CTK} = \frac{BT}{CT} \cdot \frac{\sin \angle MCK}{\sin \angle MBK} \\ &= \frac{BT}{CT} \cdot \frac{BM}{CM} = \frac{BT}{CT} \cdot \frac{KA_1}{KA_2} = \frac{BT}{CT} \cdot \frac{AP}{AQ}. \end{aligned}$$
Proof completed.

Proof 1 (Alternative). An alternative proof that points $B, C, M$, and $T$ are concyclic.
Since $K, P, Q$, and $T$ are concyclic, by Ptolemy's Theorem, $KT \cdot PQ = PK \cdot QT + QK \cdot PT$. Also, by the Law of Sines, $\frac{PQ}{\sin \angle PKQ} = \frac{TQ}{\sin \angle TKQ} = \frac{PT}{\sin \angle PKT}$, hence
$$KT = \frac{PK \cdot QT + QK \cdot PT}{PQ} = \frac{PK \cdot \sin \angle TKQ + QK \cdot \sin \angle PKT}{\sin \angle PKQ}.$$
Let $\angle ABC = \angle B$ and $\angle ACB = \angle C$.
Noting that $KP = 2BK \cos \angle B$ and $KQ = 2CK \cos \angle C$, and utilizing the parallel relationships given in the problem, we obtain
$$\begin{aligned} KT &= \frac{2BK \cos \angle B \cdot \sin \angle CAK + 2CK \cos \angle C \cdot \sin \angle BAK}{\sin(\angle B + \angle C)} \\ &= \frac{2BK \cos \angle B \cdot \frac{KC}{KA} \sin \angle C + 2CK \cos \angle C \cdot \frac{KB}{KA} \sin \angle B}{\sin(\angle B + \angle C)} \\ &= \frac{\cos \angle B \cdot \sin \angle C + \cos \angle C \cdot \sin \angle B}{\sin(\angle B + \angle C)} \cdot \frac{2KB \cdot KC}{KA} = \frac{KB \cdot KC}{KM}, \end{aligned}$$
therefore $KM \cdot KT = KB \cdot KC$. By the Power of a Point Theorem, it follows that points $B, C, M$, and $T$ are concyclic. Proof completed.

Proof 2. Let's consider the circumcircle of $\triangle ABC$ as the unit circle in the complex plane. We use uppercase letters for points and the corresponding lowercase letters for their complex numbers.
Let $X$ be the intersection point of the line segment $AK$ with the unit circle. Then
$$\bar{k} = \frac{a + x - b - c}{ax - bc}.$$
Thus,
$$k = \frac{\bar{a} + \bar{x} - \bar{b} - \bar{c}}{\bar{a}\bar{x} - \bar{b}\bar{c}} = \frac{\frac{1}{a} + \frac{1}{x} - \frac{1}{b} - \frac{1}{c}}{\frac{1}{ax} - \frac{1}{bc}} = \frac{bc(x + a) - ax(c + b)}{bc - ax} = \frac{(ax - ac - cx)b + axc}{ax - bc}.$$
Note the following geometric facts: the perpendicular bisector of $KP$ passes through $B$ and is perpendicular to $AB$, the perpendicular bisector of $KQ$ passes through $C$ and is perpendicular to $AC$. Therefore, the circumcenter of $\triangle PQK$ is the diametrically opposite point of $A$ on the unit circle, denoted as $A_1$. From $\angle AXA_1 = \frac{\pi}{2}$, we know $X$ is the midpoint of $TK$.
Therefore,
$$k = \frac{\frac{1}{a} + \frac{1}{x} - \frac{1}{b} - \frac{1}{c}}{\frac{1}{ax} - \frac{1}{bc}} = \frac{bc(x + a) - ax(c + b)}{bc - ax} = \frac{(ax - ac - cx)b + axc}{ax - bc}.$$

We observe the following geometric facts: The perpendicular bisector of the segment $KP$ is the line passing through point $B$ and perpendicular to $AB$, and the perpendicular bisector of $KQ$ is the line passing through $C$ and perpendicular to $AC$. Therefore, the circumcenter of $\triangle PQK$ is the diametrical opposite point of $A$ on the unit circle, denoted as $A_1$. Thus, according to $\angle AXA_1 = \frac{\pi}{2}$, we know that $X$ is the midpoint of $TK$.
So
$$\begin{aligned} t &= 2x - k = 2x - \frac{(ax - ac - cx)b + acx}{ax - bc} = \frac{2ax^2 - (ax - ac + cx)b - axc}{ax - bc}, \\ t - b &= \frac{2ax^2 - (ax - ac + cx)b - axc - axb + b^2c}{ax - bc} = \frac{(b - x)(bc + ac - 2ax)}{ax - bc}, \\ t - c &= \frac{2ax^2 - (ax - ac + cx)b - axc - axc + bc^2}{ax - bc} = \frac{(c - x)(ab + bc - 2ax)}{ax - bc}. \end{aligned}$$
On the other hand, since $Q$ and $K$ are symmetric about the line $CA_1$ (noting that $a_1 = -a$), we have
$$\frac{q+a}{c+a} = \frac{\overline{k+a}}{c+a} = \frac{(\bar{k}+\bar{a})ac}{a+c} = \frac{ac\bar{k}+c}{a+c},$$
$$\text{thus } q = -a + c + ac\bar{k} = (c - a) + \frac{ac(a+x-b-c)}{ax-bc}. \text{ Hence,}$$
$$\begin{aligned} q - c &= \frac{-a(ax - bc) + ac(a + x - b - c)}{ax - bc} = \frac{-a^2x + a^2c + acx - ac^2}{ax - bc} = \frac{a(a - c)(c - x)}{ax - bc}, \\ q - a &= \frac{(c - 2a)(ax - bc) + ac(a + x - b - c)}{ax - bc} = \frac{abc - bc^2 - 2a^2x + 2acx + a^2c - ac^2}{ax - bc} \\ &= \frac{(a - c)(ac + bc - 2ax)}{ax - bc}. \end{aligned}$$
$$\text{Similarly, we have } p = -a + b + ab\bar{k} = (b - a) + \frac{ab(a+x-b-c)}{ax-bc}. \text{ Thus,}$$
$$\begin{aligned} p - b &= \frac{-a(ax - bc) + ab(a + x - b - c)}{ax - bc} = \frac{-a^2x + a^2b + abx - ab^2}{ax - bc} = \frac{a(a - b)(b - x)}{ax - bc}, \\ p - a &= \frac{(b - 2a)(ax - bc) + ab(a + x - b - c)}{ax - bc} = \frac{abc - b^2c - 2a^2x + 2abx + a^2b - ab^2}{ax - bc} \\ &= \frac{(a - b)(ab + bc - 2ax)}{ax - bc}. \end{aligned}$$
Therefore, we arrive at
$$\frac{t-c}{t-b} \cdot \frac{p-b}{p-a} = \frac{q-c}{q-a},$$
and by taking the argument of both sides of this equation, we obtain conclusion (1); by taking the modulus, we obtain conclusion (2).