They are visible because they cast shadows onto the retina, either by completely occluding the path of light or via refractive means. Imagine a dark box forming a camera obscura. If we place an object in the box, we will form a sort of image on the wall of the box by virtue of casting a shadow on it.
Your second and third questions are really the same. It is the same reason that decreasing the aperture of a lens will automatically increase the depth-of-field. A narrower aperture restricts the range of angles by which a ray can be transported from the object to the image plane. Consider that each point on the object/floater will be transported to a variety of locations on the retina, depending on the way that the particular ray enters the pupil. Narrowing the pupil decreases the size of this region on the retina, leading to a sharper image. See this poor diagram I drew:
$A$ represents the pupil, $B$ the floater, and $C$ the retina. You can see that the narrower pupil restricts the available paths mapping an incoming ray through a given point on the floater to the retina.
The inverse is the reason they become invisible upon dilation of the pupil - the pupil becomes sufficiently large that a given target location on the retina receives light from distant parts of the object, preventing formation of a useful image. Consider in our camera obscura analogy how narrowing the pinhole would bring the shadow-image into sharper view, whereas increasing the size of the pinhole would eventually render the shadow invisible.
