If you have a parametric equation $x=x(t)$ and $y=y(t)$, then
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.$$
In your case, you have $x(\theta)=r\theta-r\sin\theta$ and $y(\theta)=r-r\cos\theta$. Therefore, $\dfrac{dx}{d\theta} = \ldots$ and $\dfrac{dy}{d\theta} = \ldots$, and hence
$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \ldots$$
Then you'll want to compute $\left.\dfrac{dy}{dx}\right|_{\theta=\pi/6}$ and $\left.\dfrac{dy}{dx}\right|_{\theta=5\pi/4}$ to get the slopes of the two tangent lines. Once you have those, you'll be able to compute the tangent line equations like usual using point-slope form; in particular, your two tangent line equations will be
$$y=\left(\left.\dfrac{dy}{dx}\right|_{\theta=\pi/6}\right)(x-x(\pi/6)) + y(\pi/6)$$
and
$$y=\left(\left.\dfrac{dy}{dx}\right|_{\theta=5\pi/4}\right)(x-x(5\pi/4)) + y(5\pi/4)$$
To see where the tangent line is horizontal, you'll want to solve $\dfrac{dy}{d\theta}=0$.
To see where the tangent line is vertical, you'll want to solve $\dfrac{dx}{d\theta}=0$.
To see if the cycloid is concave up or down, you'll need to compute the second derivative:
$$\frac{d^2y}{dx^2} = \dfrac{d}{dx}\left(\frac{dy}{dx}\right) = \dfrac{\dfrac{d}{d\theta}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{d\theta}}$$
and see where it's positive and/or negative.
I hope this provides you with enough information to at least start the problem.
This seems to be the question that I answered earlier. $x_0 = r\cos \theta = (3 - 3\sin 3\pi/4)\cos 3\pi/4$, and similarly you can find $y_0$, and then use $y - y_0 = m(x - x_0)$ with $m = 2\sqrt{2}-3$
Best Answer
$\cos\theta\cos(3\theta)-3\sin(3\theta)\sin\theta=1/2(\cos2\theta+\cos4\theta-3(\cos2\theta-\cos4\theta))=0\implies \cos2\theta=2\cos4\theta$.$$$$ Now $\cos4\theta=2\cos^2 2\theta-1$.Putting this in equation and letting $\cos2\theta =t$ gives, $$t=2(2t^2-1)$$ $$\implies 4t^2-t-2=0$$ $$\implies t=\frac{1\pm \sqrt{33}}{8}$$ and then $\theta=\frac{\cos^{-1}(t)}{2}$ and then you can find the equation easily.