All of the question are in sequence and related.
1.Given 2 ellipse with the position x1,y1, x2,y2 and the radius a1,b1, a2,b2, construct an equation to determine if both of them has at least one intersection point (note that the point of intersection is not important). Assume the angle of rotation is 0 (not rotated).
2.Adds theta1 and theta2 to represent the angle of rotation of the ellipses.
3.Beside checking for intersection, reconstruct the equation to also check if one of the ellipse is inside another ellipse.
Rotate your ellipses so that the first of them has the axes aligned to the coordinate axes;
Dilate along one of the axis (i.e. the $x$-axis) so that the first ellipse becomes a circle;
Rotate the other ellipse around the origin so that its axes become aligned to the coordinate axes.
With these transformation you reduce your problem to the case when the first ellipse is a circle centered in the origin and the other is axis aligned.
To know if the circle and the ellipse do intersect you can find the point of the ellipse which is closest to the origin (and hence to the circle) which is a point $(x,y)$ such that the normal at $(x,y)$ to the ellipse points exactly towards the origin.
Mathematically this becomes an equation of third order, which is not good to solve algebraically. In these cases the best option is to proceed numerically. If the ellipse has equation $$ x(t) = x_0 + \alpha \cos t\\ y(t) = y_0 + \beta \sin t $$ it can be written as $$ \frac{(x-x_0)^2}{\alpha^2} + \frac{(y-y_0)^2}{\beta^2} = 1 $$ and its normal vector at a point $(x,y)$ has the direction of the gradient of the previous function: $$ \nu(x,y) = 2\left(\frac{x-x_0}{\alpha^2}, \frac{y-y_0}{\beta^2}\right). $$ Imposing $\nu$ being parallel to $(x,y)$ means that: $$ \beta^2 y (x-x_0) - \alpha^2 x(y-y_0) = 0 $$ which becomes an equation in $t$: $$ \beta(y_0+\beta \sin t) \cos t - \alpha (x_0+\alpha \cos t) \sin t = 0. $$
This equation can be solved numerically with very efficient algorithms. For example one can use bisection method. Remember that we expect two solutions to this equation (the minimum and maximum distance points) so we need to choose the right starting interval in which $t$ should vary. The right interval would be one of the four quadrants: $[0,\pi/2]$, $[\pi/2,\pi]$, $[-\pi,-\pi/2]$, $[-\pi/2,0]$ corresponding to the opposite of the quadrant where the center of the ellipse is located.
Then the bisection method quickly converges to the desired point.