1) Let the center of the circle be $I$. Draw $IM$ and $QN$ perpendicular to the $x-$axis and intersect the $x-$axis at $M$ and $N$. Draw $IK$ perpendicular to $QN$ and intersects the latter at $K$. Then $$\angle{IQK}=90^{\circ}-\angle{NQP}=\angle{QPN}=\theta$$

It follows that $$ON = OM + MN = OM + IK = OM + IQ\sin\theta = 1 + \sin\theta$$

and $$QN = KN + QK = KN + IQ\cos\theta = 1 + \cos\theta$$

This means the coordinates of $Q$ is $(1+\sin\theta, 1 + \cos\theta)$.

Now the line $PR$ has a slope of $\tan(\pi-\theta)=-\tan\theta$ and passes point $(1+\sin\theta, 1+\cos\theta)$. Hence, its equation is $$y-(1+\cos\theta)=-\tan\theta(x-(1+\sin\theta))\implies y=-\tan\theta x +\tan\theta(1+\sin\theta)+(1+\cos\theta)$$

Setting $y=0$ leads $$x=1 + \sin\theta + \frac{1+\cos\theta}{\tan\theta}=\frac{\sin\theta + \sin^2\theta + \cos\theta + \cos^2\theta}{\sin\theta}=1+\csc\theta+\cot\theta$$

2) When $\angle{RPO}=\theta$, $\angle{PRO}=\frac{\pi}{2}-\theta$. Imagine that $\angle{RPO}$ becomes $\frac{\pi}{2}-\theta$, then $\angle{PRO}$ becomes $\theta$. At this time, the diagram is just a reflection of the original one: the current area represented by $A(\theta)$ becomes $B(\theta)$ and the current area presented by $B(\theta)$ becomes $A(\theta)$. Therefore, by the principle of symmetry, we must have $A(\theta)=B\left(\frac{\pi}{2} - \theta\right)$.

 

When $\theta=\frac{\pi}{4}$, then $A(\theta)=B(\theta)$. In this case, $$OP = OR = 1+\csc 45^{\circ} + \cot 45^{\circ} =2+\sqrt{2}\implies S_{\triangle{PRO}}=\frac{1}{2}\left(2+\sqrt{2}\right)^2=3+2\sqrt{2}$$

Meanwhile, the area of the circle is $\pi$ and the area in the left-bottom corner equals $\left(1 -\frac{\pi}{4}\right)$. Therefore $$A\left(\frac{\pi}{4}\right)=\frac{1}{2}\left(\left(3+2\sqrt{2}\right) - \pi - \left(1 -\frac{\pi}{4}\right)\right)=\boxed{1+\sqrt{2}-\frac{3\pi}{8}}$$

3) When $\theta=\frac{\pi}{3}$, then $\angle{QIM}=\frac{2\pi}{3}$. This means that the to be determined area equals the area of $PMIQ$ minus two third of the area of the circle. Meanwhile, the area of $PMIQ$ is the sum of two congruent right triangles $PMI$ and $PQI$. Hence $$A\left(\frac{\pi}{3}\right)=2\times \frac{1}{2}\left(\left(1+\frac{2\sqrt{3}}{3} + \frac{\sqrt{3}}{3}\right)-1\right)\times 1 - \frac{\pi}{3} = \boxed{\sqrt{3}-\frac{\pi}{3}}$$