[list=a][*] Find infinitely many pairs of integers $a$ and $b$ with $1<a<b$, so that $ab$ exactly divides $a^{2}+b^{2}-1$. [*] With $a$ and $b$ as above, what are the possible values of \[\frac{a^{2}+b^{2}-1}{ab}?\] [/list]
**
   The equation \(a^2 + b^2 - 1 = kab\) is symmetric in \(a\) and \(b\). We start with the known solution \((a, b) = (2, 3)\). For this pair:
   \[
   a^2 + b^2 - 1 = 2^2 + 3^2 - 1 = 4 + 9 - 1 = 12
   \]
   \[
   ab = 2 \cdot 3 = 6
   \]
   \[
   \frac{a^2 + b^2 - 1}{ab} = \frac{12}{6} = 2
   \]
   Thus, \(k = 2\).

2. **Vieta Jumping:**
   Using Vieta's formulas, for the quadratic equation \(a^2 - kab + b^2 - 1 = 0\), the roots \(a_1\) and \(a_2\) satisfy:
   \[
   a_1 + a_2 = kb \quad \text{and} \quad a_1 a_2 = b^2 - 1
   \]
   Given \(a_1 = a\) and \(a_2 > b\) for \(k > 1\), we can generate new pairs \((a, b)\) by swapping \(a\) and \(b\) and applying the same process iteratively.

3. **Generating New Pairs:**
   By changing \(a\) and \(b\) and using the symmetry, we can generate infinitely many pairs \((a, b)\). For example, starting from \((2, 3)\), we can find new pairs by solving the quadratic equation for new values of \(a\) and \(b\).

### Part (b)
We need to show that all positive numbers greater than 1 are possible values of \(\frac{a^2 + b^2 - 1}{ab}\).

1. **Setting \(a = k\) and \(b = k^2 - 1\):**
   Let \(a = k\) and \(b = k^2 - 1\) where \(k \in \mathbb{N}\) and \(k > 1\). Then:
   \[
   a^2 + b^2 - 1 = k^2 + (k^2 - 1)^2 - 1
   \]
   \[
   = k^2 + (k^4 - 2k^2 + 1) - 1
   \]
   \[
   = k^2 + k^4 - 2k^2
   \]
   \[
   = k^4 - k^2
   \]

2. **Calculating the Ratio:**
   \[
   \frac{a^2 + b^2 - 1}{ab} = \frac{k^4 - k^2}{k(k^2 - 1)}
   \]
   \[
   = \frac{k^2(k^2 - 1)}{k(k^2 - 1)}
   \]
   \[
   = k
   \]
   Since \(a = k\) and \(b = k^2 - 1\) are both greater than 1, and \(a^2 + b^2 - 1 > ab\), we get \(k > 1\).

Thus, all positive integers greater than 1 are possible values of \(\frac{a^2 + b^2 - 1}{ab}\).

The final answer is \( \boxed{ k } \) where \(k > 1\).