(a) Prove that given constants $a,b$ with $1<a<2<b$, there is no partition of the set of positive integers into two subsets $A_0$ and $A_1$ such that: if $j \in \{0,1\}$ and $m,n$ are in $A_j$, then either $n/m <a$ or $n/m>b$.
(b) Find all pairs of real numbers $(a,b)$ with $1<a<2<b$ for which the following property holds: there exists a partition of the set of positive integers into three subsets $A_0, A_1, A_2$ such that if $j \in \{0,1,2\}$ and $m,n$ are in $A_j$, then either $n/m <a$ or $n/m>b$.
### Part (a)

1. **Assume a partition exists:**
   Suppose there exists a partition of the set of positive integers into two subsets \( A_0 \) and \( A_1 \) such that if \( j \in \{0,1\} \) and \( m, n \) are in \( A_j \), then either \( \frac{n}{m} < a \) or \( \frac{n}{m} > b \).

2. **Coloring scheme:**
   Color a positive integer \( x \) white if it's in \( A_0 \) and black if it's in \( A_1 \).

3. **Select a large \( x \):**
   Select some large \( x \in \mathbb{N} \). Without loss of generality, assume \( x \) is black (i.e., \( x \in A_1 \)).

4. **Intervals and functions:**
   Define \( f(n) \) as the smallest integer greater than \( an \) and \( g(n) \) as the greatest integer less than \( bn \). For all \( n \in \mathbb{N} \), we know that \( [f^{(k)}(x), g^{(k)}(x)] \) is all the same color for sufficiently large \( k \) and \( x \).

5. **Contradiction:**
   For sufficiently large \( k \) and \( x \), there exist some two numbers \( y, z \) in the interval \( [f^{(k)}(x), g^{(k)}(x)] \) such that \( \frac{z}{y} \in [a, b] \). This contradicts the assumption that \( \frac{n}{m} < a \) or \( \frac{n}{m} > b \) for all \( m, n \) in the same subset \( A_j \).

Thus, no such partition exists.

\(\blacksquare\)

### Part (b)

1. **Claim:**
   We claim that the answer is all real numbers \( (a, b) \) satisfying \( 1 < a < 2 < b \) and \( b \le a^2 \).

2. **Constructing the partition:**
   To show that these pairs work, partition the positive integers into three subsets \( A_0, A_1, A_2 \) such that a positive integer \( n \) is in \( A_i \) if \( \left\lfloor \log_a (n) \right\rfloor \equiv i \pmod{3} \).

3. **Verification:**
   If \( m, n \) are in the same subset \( A_j \), then \( \left\lfloor \log_a (m) \right\rfloor \equiv \left\lfloor \log_a (n) \right\rfloor \pmod{3} \). This implies that \( \frac{n}{m} \) is either less than \( a \) or greater than \( a^2 \). Since \( b \le a^2 \), it follows that \( \frac{n}{m} \) is either less than \( a \) or greater than \( b \).

4. **Failure for \( b > a^2 \):**
   Suppose \( b > a^2 \). Consider a large enough \( x \). Since \( x \) is large and \( b > a^2 \), the numbers \( x, f(x), \) and \( g(x) \) must be in different sets. This also holds if \( g(x) \) is replaced by any number in the interval \( [f(f(x)), g(x)] \). Hence, for sufficiently large \( x \), the entire interval \( [f(f(x)), g(x)] \) is in the same set. By applying this for \( x, x+1, x+2, \ldots \), we can obtain that all sufficiently large numbers are in the same set, which is a contradiction.

Thus, the pairs \( (a, b) \) satisfying \( 1 < a < 2 < b \) and \( b \le a^2 \) are the only ones for which the partition exists.

\(\blacksquare\)

The final answer is \( \boxed{ (a, b) } \) satisfying \( 1 < a < 2 < b \) and \( b \le a^2 \).