There was lemniscate trouble, as you feared. Up to $\theta=\frac{\pi}{4}$, everything was fine. But from $\frac{\pi}{4}$ and for quite a while (up to $\frac{3\pi}{4}$), $\cos 2\theta$ is negative, so $r^2$ is negative and there is no curve. The integral doesn't know and doesn't care: it cheerfully "adds up" these negatives, giving the wrong answer.
So for the limaçon part, if you do things in your style, you will have to break up the limaçon integral into two integrals, $\frac{\pi}{6}$ to $\frac{\pi}{4}$ and then $\frac{3\pi}{4}$ to $\frac{5\pi}{6}$.
Since the integrals for the lemniscate and the limaçon are over different intervals, we cannot express their difference as a single integral.
What I would do is to use the symmetry, take the right-half of the region and multiply the resulting area by $2$. The limaçon part to be subtracted uses the integral from $\frac{\pi}{6}$ to $\frac{\pi}{4}$.
So there is $\frac{1}{2}$ as in your formula, but ultimately we multiply by $2$, so our area is
$$\int_{\pi/6}^{\pi/2}(3+2\sin \theta)^2\,d\theta-\int_{\pi/6}^{\pi/4}32\cos 2\theta\,d\theta.$$
More pleasant, fewer minus signs to worry about!
You need not just solve for $f(\theta) = g(\theta)$, but also solve for $f(\theta) = -g(\pi + \theta)$, to take into account the chance that a point of intersection of this form.
For your example $f = 1+\cos\theta, g = 1-\cos\theta$, so you need to solve for $1+\cos\theta = -1+\cos(\theta+\pi) = -1-\cos\theta$. This simplifies to $\cos\theta = -1$, so $\theta = \pi$, as desired.
edit: I just realized that you generally need to check the origin as a special case, too, since $r,\theta$ represents the origin whenever $r=0$ no matter the value of $\theta$.
Best Answer
I find your question interesting.
The trick is that the pole has coordinates $(0,\theta)$, no matter what $\theta$ you choose. So solving $1+\cos\theta = 3\cos \theta$ to find $\theta$ is not necessarily relevant to find this point.
One way of doing it is by asking yourself if for each curve, there is an angle $\theta$ for which $r(\theta)=0$. Clearly it is the case: $\theta_1=\pi/2$ for $r=3\cos\theta$, and $\theta_2 = \pi$ for $r=1+\cos\theta$.
So you have proved that each curve will cross the pole at least once, therefore it is indeed an intersection point of the curves.