Adding a few more perpendiculars, we augment $PA$ to a rectangle $\square PAQA^\prime$ with diagonal $PQ$ a diameter of the smaller circle. (We've exploited the fact that any angle inscribed in a semicircle is a right angle.) This makes $\square AQBC$ an isosceles trapezoid, whose height we'll denote $q$, smaller base $p$, and larger base $p+2s$.
Note that Pythagoras gives us
$$|PQ|^2 = p^2 + q^2 = 4 r^2 \qquad |CA|^2 = q^2 + s^2 \qquad |AB|^2 = q^2 + (p+s)^2$$
Moreover, the (unsigned) Power of Point $P$ relative to the big circle is
$$R^2 - r^2 = |PC| |PB| = s \left( p + s \right)$$
|AB|^2 + |BC|^2 + |CA|^2 &= \left( q^2 + \left( p + s \right)^2 \right) + \left( p + 2 s \right)^2 + \left( q^2 + s^2 \right) \\
&= 2 \left( p^2 + q^2 \right) + 6 s \left( p + s \right) \\
&= 8 r^2 + 6 \left( R^2 - r^2 \right) \\
&= 2 \left( 3 R^2 + r^2 \right)
(Similar algebra ---and an identical result--- arises from applying Ptolemy's theorem to the (necessarily-cyclic) isosceles trapezoid:
$$|CA| |BQ| + |AQ| |BC| = |AB||QC|$$
where $|BQ| = |CA|$ and $|QC| = |AB|$.)
Consequently, the sum-of-squares is independent of the location of point $B$, so that the answer to (i) is the singleton set containing $2\left( 3R^2 + r^2\right)$.
For (ii), extend the trapezoid's shorter base to match the longer, obtaining rectangle $\square BCC^\prime B^\prime$.
The midpoint, $M$, of $AB$ is always the midpoint of $PB^\prime$, whereas $B^\prime$ is always a point on the larger circle. Thus, $M$ is a dilation of $B^\prime$ in $P$ with scale factor $1/2$, and the locus of $M$ is the corresponding dilation of the locus of $B^\prime$ (aka, the big circle).
Note: The locus of $N$, the midpoint of $AC$ (and of $PC^\prime$), is the same circle. Likewise with the midpoints of $PC$ and $PB$.