Solution:

a.
For convenience, we assume the $a_{i}$ are indexed modulo $101$, so that $a_{i+1} = a_{1}$ when $a_{i} = a_{101}$.

In any consistent subset of $C_{101}$ of order $1$, $b$ must be paired with exactly one $a_{i}$, say $a_{1}$. Then, $a_{2}$ cannot be paired with $a_{1}$, so it must be paired with $a_{3}$, and likewise we find we use the pairs $\{a_{4}, a_{5}\}, \{a_{6}, a_{7}\}, \ldots, \{a_{100}, a_{101}\}$—and this does give us a consistent subset of order 1. Similarly, pairing $b$ with any other $a_{i}$ would give us a unique extension to a consistent configuration of order 1. Thus, we have one such 2-configuration for each $i$, giving $S_{1}(101) = 101$ altogether.

In a consistent subset of order $2$, $b$ must be paired with two other elements. Suppose one of them is $a_{i}$. Then $a_{i}$ is also paired with either $a_{i-1}$ or $a_{i+1}$, say $a_{i+1}$. But then $a_{i-1}$ needs to be paired up with two other elements, and $a_{i}$ is not available, so it must be paired with $a_{i-2}$ and $b$. Now $b$ has its two pairs determined, so nothing else can be paired with $b$. Thus, for $j \neq i-1, i$, we have that $a_{j}$ must be paired with $a_{j-1}$ and $a_{j+1}$. So our subset must be of the form
$$\{\{b, a_{i}\}, \{a_{i}, a_{i+1}\}, \{a_{i+1}, a_{i+2}\}, \ldots, \{a_{101}, a_{1}\}, \ldots, \{a_{i-2}, a_{i-1}\}, \{a_{i-1}, b\}\}$$
for some $i$. On the other hand, for any $i = 1, \ldots, 101$, this gives a subset meeting our requirements. So, we have $101$ possibilities, and $S_{2}(101) = 101$.

Finally, in a consistent subset of order $3$, each $a_{i}$ must be paired with $a_{i-1}, a_{i+1}$, and $b$. But then $b$ occurs in $101$ pairs, not just $3$, so we have a contradiction. Thus, no such subset exists, so $S_{3}(101) = 0$.

b.
No more than two of the pairs $\{v, x\}, \{v, y\}, \{w, y\}, \{w, z\}, \{x, z\}$ may be included in a 2-configuration of order 1, since otherwise at least one of $v, w, x, y, z$ would occur more than once. If exactly one is included, say $\{v, x\}$, then $w, y, z$ must be paired with $W, Y, Z$, respectively, and then $V$ and $X$ cannot be paired. So either none or exactly two of the five pairs above must be used. If none, then $v, w, x, y, z$ must be paired with $V, W, X, Y, Z$, respectively, and we have $1$ 2-configuration arising in this manner. If exactly two are used, we can check that there are $5$ ways to do this without duplicating an element:
$$\{v, x\},\{w, y\} \quad \{v, x\},\{w, z\} \quad \{v, y\},\{w, z\} \quad \{v, y\},\{x, z\} \quad \{w, y\},\{x, z\}$$
In each case, it is straightforward to check that there is a unique way of pairing up the remaining elements of $A$. So we get $5$ 2-configurations in this way, and the total is $6$.

c.
Let $A_{n} = \{a_{1}, b_{1}, a_{2}, b_{2}, \ldots, a_{n}, b_{n}\}$ for $n \geq 1$, and consider the 2-configuration $C_{n}$ consisting of $\{a_{i}, b_{i}\}$ for all $1 \leq i \leq n$, $\{a_{i}, a_{i+1}\}$ for all $1 \leq i \leq n-1$, and $\{b_{i}, b_{i+1}\}$ for all $1 \leq i \leq n-1$. Let $N_{n}$ be the number of subsets of $C_{n}$ that are consistent of order 1 (call these "matchings" of $C_{n}$). Consider any matching of $C_{n+2}$. Either $a_{n+2}$ is paired with $b_{n+2}$, in which case the remaining elements of our matching form a matching of $C_{n+1}$; or $a_{n+2}$ is paired with $a_{n+1}$, in which case $b_{n+2}$ must be paired with $b_{n+1}$, and the remaining elements form a matching of $C_{n}$. It follows that $N_{n+2} = N_{n+1} + N_{n}$. By direct calculation, $N_{1} = 1$ and $N_{2} = 2$, and now computing successive values of $N_{n}$ using the recurrence yields $N_{10} = 89$.