Here is the solution to problem 2:
Given that \( b \) is the inverse of \( a \) modulo 44, we have the relationship:
\[
ab \equiv 1 \pmod{44}
\]
We need to find the inverse of \( 9a \) modulo 44 in terms of \( b \). Let the inverse of \( 9a \) modulo 44 be \( x \). This means:
\[
(9a)x \equiv 1 \pmod{44}
\]
We can factor out the \( a \):
\[
9(ax) \equiv 1 \pmod{44}
\]
Since \( ab \equiv 1 \pmod{44} \), we replace \( ax \) with \( b \):
\[
9b \equiv 1 \pmod{44}
\]
Thus, \( x = 9b \) is the inverse of \( 9a \) modulo 44.
So, the inverse of \( 9a \) modulo 44 in terms of \( b \) is \( \boxed{9b} \).
