$f(r,\theta)=(r\cos\theta,r\sin\theta)$ Then $f(r,\theta)=(r\cos\theta,r\sin\theta)$ Then
$1.$ $Df(r,\theta)$ is always nonzero for any $(r,\theta)\in\mathbb{R}^2,r\ne 0$
$2.$ $f$ is injective on $\{(r,\theta):r\ne 0\}$
$3.$ On any neighborhood of $(r,\theta):r\ne 0$, $f$ is one one.
$4$ $Df(r,\theta)=r^2I\forall (r,\theta):r\ne 0$
I have calculated that $Df(r,\theta)$ is the matrix $$\begin{bmatrix} \cos\theta &-r \sin\theta \\ \sin\theta & r\cos\theta \end{bmatrix}$$ so its determinant is r which is nonzero.so by inverse mapping theorem $3$ is true, $1$ is true also, I don't know about $2$ and $4$, could any one tell me how to prove or disprove , for $4$ I have taken $(r,\theta)=(r,\pi)$ and saw the contradiction.

For 4, you have already set up a counterexample (provided you chose $r\neq 0$).
Hint: For 2 note that $\sin$ and $\cos$ are both periodic with period $2\pi$. So it is pretty easy to find $\theta$ and $\theta'$ with $f(r,\theta)=f(r,\theta')$. Therefore 2 is false.
