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
$l=r+r+(2\times\pi\times r)=2(r+\pi)$

2) If the two edges are the same
length, how long do they need to be to ensure every point on the
circumference touches? What if they're not the same
length?

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?

Initially just looking at one side of the tray we can see how
much of the circumference touches this. We can find the arc length
that touches on any side by looking at the length of the side minus
two radii from the corners. By noting the formula $c\times f=a$,
the formula relating the cicumference, fraction of the
circumference and the arc length of that fraction, we find
$f=\frac{l-2r}{2\pi r}$.

So for a disc of radius $1$ in the $4 \times 3$ box, the
proportion touched on the first side is $\frac{4-2r}{2\pi
r}=\frac{1}{\pi}$, and then a quarter is missed, then
$\frac{3-2r}{2\pi r}=\frac{1}{2\pi}$ and miss a quarter again. Then
repeat for the last two sides. So after two sides
$\frac{1}{\pi}+\frac{1}{2\pi}=\frac{3}{2\pi}$ of the circumference
has been touched, and we have moved $\frac{3}{2\pi}+\frac{1}{2}$ of
the circumference. Which is a distance of $3+\pi$, just less than
the actual circumference of $2\pi$. So then the new bits that will
be touched are $2\pi-(3+\pi)=\pi-3$ on each side from now on. So on
the last two sides $2(\pi-3)$ of the circumfernce is touched, which
is $\frac{2(\pi-3)}{2\pi}=\frac{\pi-3}{\pi}$ of the circumference.
Adding this to the proportion touched on the first two sdies, the
total proportion touched is
$\frac{3}{2\pi}+\frac{\pi-3}{\pi}=\frac{2\pi-3}{2\pi}$ of the
circumfernce. Which is a total arc length of $2\pi \times
\frac{2\pi-3}{2\pi}=2\pi-3$.

Then as for every side the disc touches it touches anew
$\pi-3$ of the circumference it will eventually touch the entire
circumfernce.