$r=1-\sin(\theta)$ horizontal and vertical tangents
Vertical tangents: Your values of $\theta$ are correct, so we can find the $x$ and $y$ coordinates of the intersections by just plugging into $x=\cos\theta(1-\sin\theta)$ and $y=\sin\theta(1-\sin\theta)$. That gives $$ (x,y) = (\pm \frac{3\sqrt{3}}{4},-\frac{3}{4}) $$ So the tangent lines are given by $x=\pm \frac{3\sqrt{3}}{4}$, and they intersect the curve at $y=-\frac34$.
Is this what you wanted?