Contact

Well done Sam from Kings Wimbledon for your solution:

1) Coin radius 1: How long would the box need to be for every point on the coin's circumference to touch the box as I roll it along?

The left bottom quarter of the disc is not in contact, which is a radius' width. Then the whole circumference must roll along the base of the box, until another quarter of circumference is not in contact in the bottom right and corner.
So the disc must be two radii (from when the disc is at the sides) plus one circumference long. This is $l=1+1+(2\times\pi\times1)=2(1+\pi)=8.28$.
For any radius $r$, the length of the box is $l=r+r+(2\times\pi\times r)=2(r+\pi)$

Call the length of the edges $l$. Then the amount of the circumference that touches each edge is $l-r$, as at the corner there will be a quarter of the circumference that will not be touched.
Obviously if $l-r> c$ then the whole circumference will touch both edges, so we want to find the minimum length an edge can have to touch the whole circumference. So you need
$2\times(l-r)+\frac{2\pi r}{4}=2\times(l-r)+\frac{\pi r}{2}=2\times(2\pi r)$
$2\times(l-r)=\frac{7}{2} \times \pi r$
$l=\frac{7}{4} \times \pi r +r$
$l=r\times(\frac{7}{4}\pi + 1)$
That is, $\frac{7}{8}$ of the circumference, and then plus the extra radius that is missed in the corner.

For edges of different lengths, obviously if both lengths are at least as long as $r\times(\frac{7}{4}\pi + 1)$ then it will work, and the whole circumference will be covered.
But the more interesting question is, if one of the edges is shorter than this figure, how long will the other need to be to make up for it, and ensure the whole circumference is still touched.
The answer is that if one length is less than $r\times(\frac{7}{4}\pi + 1)$, then the other length must be the entire circumference plus the extra radius lost in the corner. This can be seen from labeling four equidistant points on the disc and studying their movement as the disc rotates around the edges.

3) Rectangular tray: What proportion of the circumference touches the tray as it makes one complete circuit of a 4 by 3 tray?