1. Consider two concentric circles on the same plane with radii $R$ and $r (R > r)$. Let $P$ be a fixed point on the circumference of the smaller circle, and $B$ be a moving point on the circumference of the larger circle. The line $B P$ intersects the larger circle at another point $C$. The line $l$ passing through point $P$ and perpendicular to $B P$ intersects the smaller circle at another point $A$ (if $l$ is tangent to the smaller circle at $P$, then $A = P$).
(i) Find the set of values of the expression $B C^{2} + C A^{2} + A B^{2}$.
(ii) Determine the locus of the midpoint of line $A B$.
1. Let the centers of the two circles be $O$, and draw $OM$ perpendicular to $BC$, intersecting $BC$ at $M$. Let $AP=2t$, so $OM=t$. Thus,
$$
\begin{array}{l}
BC = 2BM = 2\sqrt{R^2 - t^2}, \\
BP^2 + CP^2 \\
= (BM - PM)^2 + (BM + PM)^2 \\
= 2(BM^2 + PM^2),
\end{array}
$$

which means $BP^2 + CP^2 = 2(R^2 - t^2) + 2(r^2 - t^2)$
$$
= 2(R^2 + r^2) - 4t^2.
$$

Therefore,
$$
\begin{array}{l}
BC^2 + CA^2 + AB^2 = BC^2 + (PC^2 + PA^2) \\
+ (BP^2 + PA^2) = BC^2 + BP^2 + CP^2 + 2PA^2 \\
= 4(R^2 - t^2) + 2(R^2 + r^2) - 4t^2 + 2PA^2 \\
= 6R^2 + 2r^2 - 8t^2 + 2(2t)^2 \\
= 6R^2 + 2r^2.
\end{array}
$$

This is a constant, independent of the position of $B$ (even in the special case where $A = P$, the expression $BC^2 + CA^2 + AB^2$ still takes the same value). Therefore, the set of values for the expression is $\{6R^2 + 2r^2\}$. This answers the first part of the problem.

Now, let's solve the second part. For $B'$, it is easy to see that $PB'A$ forms a rectangle, so the midpoint of segment $AB$ is also the midpoint of segment $PB'$. As $B$ moves around the circumference of the larger circle once, $B'$ also moves around the circumference of the larger circle once. This indicates that the locus is a circle with the midpoint of segment $OP$ as its center and $R/2$ as its radius.