Why is $r= \cos\left(\dfrac{\theta}{2}\right)$ symmetric about the Y axis?
A curve is symmetric about y axis if $(r,θ)=(−r,−θ)$ or $(r,θ)=(r,π−θ)$
But none of these is true for the above curve as $\cos\left(\dfrac{\pi- \theta}2\right) = \sin (\theta/2)$
I have realised it's symmetric about X axis as $(r,\theta ) = (r,-\theta)$.
Another problem I am facing while graphing is that how do I recognize that there's an inner part in the curve (like a loop)? What's the best way to graph this curve on XY plane?
Unlike many common examples of polar functions, the period of your function is not $2\pi$ but $4\pi$. As a result, it makes sense to consider the transformation $\theta \to 2\pi-\theta$. Note that $$ \begin{align} &\cos\left(\frac{2\pi-\theta}{2}\right) = -\cos(\theta/2), \\ &\cos(2\pi-\theta) = \cos(\theta), \text{ and } \\ &\sin(2\pi-\theta) = -\sin(\theta). \end{align} $$ As a result, $$ \begin{align} \cos\left(\frac{2\pi-\theta}{2}\right)&(\cos(2\pi-\theta),\sin(2\pi-\theta)) \\ &= -\cos(\theta/2)(\cos(\theta), -\sin(\theta)) = \cos(\theta/2)(-\cos(\theta), \sin(\theta)). \end{align} $$
Here's a modification of your Desmos graph to illustrate: