Kerr solution to Einstein field equations

I am attending a course in General Relativity and I am having some difficulties understanding the Kerr solution of Einstein field equations. In Boyer-Lindquist coordinates, this solution is written as: \begin{equation*} ds^2=-\left(1-\frac{2mr}{\Sigma}\right)dt^2-\frac{4mar \sin^2\theta}{\Sigma}dt d\varphi +\frac{\Sigma}{r^2-2ma+a^2}dr^2+\Sigma d\theta^2+\left(r^2+a^2+\frac{2ma^2r \sin^2\theta}{\Sigma}\right)\sin^2\theta d\varphi^2 \end{equation*} where \begin{equation*} \Sigma=r^2+a^2\cos^2\theta \end{equation*} and $a$ is a real parameter.

We can observe that if $a=0$, we re-obtain the better known Schwarzschild solution \begin{equation*} ds^2=-\left(1-\frac{2m}{r}\right)dt^2+\frac{1}{1-\frac{2m}{r}}dr^2+r^2\left(d\theta^2+\sin^2\theta d\varphi^2\right). \end{equation*} We also observe that, since the metric coefficients do not depend on time $t$, the Kerr solution is stationary and, similarly, since the metric coefficients do not depend on $\varphi$, the Kerr solution is axially symmetric. In other words, $\partial_t$ and $\partial_{\varphi}$ are Killing vector fields. My professor said that, unlike the Schwarzschild solution, Kerr solution is not static and not spherically symmetric, but actually I do not understand what “static solution" means. It means that $\partial_t$ is timelike ($g(\partial_t,\partial_t)<0)$? Or is it something different?

About the axial symmetry and the spherical symmetry...I cannot see the difference between Schwarzschild solution and Kerr solution: in both cases, the metric coefficients do not depend on $\varphi$ (this makes the axial symmetry...ok), but there are (in both cases!) coefficients which depend on $\theta$! So, why Schwarzschild solution is spherically symmetric, while Kerr solution is not? Maybe, the difference is that, in the case of Schwarzschild solution, $\theta$ appears only on the spherical part of the metric (which is spacelike), while in the Kerr solution $\theta$ appears also in front of $dt d\varphi$?

Still another doubt/curiosity: Kerr solution can be derived from the Schwarzschild metric written in the Eddington-Finkelstein coordinates $(v,r,\theta,\varphi)$ \begin{equation*} ds^2=-\left(1-\frac{2m}{r}\right)dv^2+2dv dr+r^2\left(d\theta^2 + \sin^2\theta d\varphi^2\right) \end{equation*} and applying the so called “Newman-Janis" algorithm, which consists in making writing the metric using a null tetrad, complexifying $r$ (i.e. assuming $r\in \mathbb{C}$), applying a complex coordinated change $r\to r+i a\cos\theta$, $v\to v+i a \cos\theta$, then calculating $g^{\alpha \beta}$ and finally inverting the matrix.

My question is: How can we be sure that this process gives us a new solution to the Einstein field equations in vacuum? \begin{equation*} G_{\mu \nu}=0. \end{equation*}

Note: I use the convention $(-,+,+,+)$ for the Lorentzian metric.