The completed truth tables yield the following results:
- For the first table: \( r = F \) and \( q \rightarrow (r \wedge p) = F \).
- For the second table: \( \sim b = T \), \( c \rightarrow \sim b = T \), \( b \rightarrow a = T \), and \( (c \rightarrow \sim b) \wedge (b \rightarrow a) = T \).
Thus, the final answers are:
- First table: \( r = F \), \( q \rightarrow (r \wedge p) = F \)
- Second table: \( \sim b = T \), \( c \rightarrow \sim b = T \), \( b \rightarrow a = T \), \( (c \rightarrow \sim b) \wedge (b \rightarrow a) = T \)
\[
\boxed{(r = F, q \rightarrow (r \wedge p) = F, \sim b = T, c \rightarrow \sim b = T, b \rightarrow a = T, (c \rightarrow \sim b) \wedge (b \rightarrow a) = T)}
\]