By covered I mean every point is inside or on a circle. I can explain this best in flat space where, if I take a circle and pack it with smaller circles, then it seems I can increase the radius of the inner circles until there are no gaps. It's ok for me that some of the inner circles will extend outside the circle to be covered. In that case all points are either inside 1 or 2 circles or are the intersection of 3 circles (so it seems to me). Inner circles will be surrounded by 6 circles. I can see that the radius of the inner circles is the distance to the centroid of the equilateral triangle formed by the centres of 3 pairwise overalapping circles and I can maybe work out the relation between the number of covering circles on the one hand and the ratio of the radii of covering and covered circles.
I need the analog for the surface of a sphere. In the case of circles of radius 1/4 the circumference of the sphere I would need just 2 circles and they would meet at an equator. Intuitively it seems not to be possible with 3 circles. What's the general case, i.e. for a given radius how many circles or vice versa, for just the cases where it's actually possible?
This is not the same as the Thomson problem or the Tammes problem. I don't understand if it's the same as this.
Your problem is the same as the third reference you give. The table from that reference, reproduced below, gives the minimal radius in degrees and the density, which is the ratio of the sum of areas of the individual circles to the area of the sphere. Areas in the density calculation are measured on the spherical surface.
Remarks on some solutions:
$n=3$: centers are at the vertices of an equilateral triangle which is inscribed in a great circle.
$n=4$: centers are at the vertices of a regular tetrahedron.
$n=5$: centers are at the vertices of a triangular bipyramid; the angular radius of the covering circles is the inverse tangent of $2$.
$n=6$: centers are at the vertices of a regular octahedron.
$n=8$: centers are not at the vertices of a cube; instead the polyhedron has triangular faces. This buckling of quadrilaterals and higher polygons is similar to the Tammes and Thomson problems.
$n=12$: centers are at the vertices of a regular icosahedron.
$n=14$: one might suppose that the optimal solution would have the cap centers at the corners and face centers of an octahedron (or equivalently, a cube); but that cap radius is about $36.20°$ and the table indicates a smaller radius is possible.