There are many $a\times b$ rectangular cardboard pieces ($a,b\in\mathbb{N}$ such that $a<b$). It is given that by putting such pieces together without overlapping one can make $49\times 51$ rectangle, and $99\times 101$ rectangle. Can one uniquely determine $a,b$ from this?
1. **Determine the area of the large rectangles:**
   - The area of the $49 \times 51$ rectangle is:
     \[
     49 \times 51 = 2499
     \]
   - The area of the $99 \times 101$ rectangle is:
     \[
     99 \times 101 = 9999
     \]

2. **Factorize the areas:**
   - Factorize $2499$:
     \[
     2499 = 49 \times 51 = 7^2 \times 3 \times 17
     \]
   - Factorize $9999$:
     \[
     9999 = 99 \times 101 = 3^2 \times 11 \times 101
     \]

3. **Find the greatest common divisor (GCD) of the two areas:**
   - The prime factorizations are:
     \[
     2499 = 3 \times 7^2 \times 17
     \]
     \[
     9999 = 3^2 \times 11 \times 101
     \]
   - The common prime factors are $3$. Therefore, the GCD is:
     \[
     \gcd(2499, 9999) = 3
     \]

4. **Determine the possible values of $a$ and $b$:**
   - Since $ab$ must divide the GCD, $ab$ must be a divisor of $3$.
   - The divisors of $3$ are $1$ and $3$.
   - Given that $a < b$ and both $a$ and $b$ are natural numbers, the only possible pair is:
     \[
     (a, b) = (1, 3)
     \]

5. **Verify the solution:**
   - Check if $a = 1$ and $b = 3$ can form the given large rectangles:
     - For the $49 \times 51$ rectangle:
       \[
       \frac{49 \times 51}{1 \times 3} = \frac{2499}{3} = 833
       \]
       This is an integer, so it is possible.
     - For the $99 \times 101$ rectangle:
       \[
       \frac{99 \times 101}{1 \times 3} = \frac{9999}{3} = 3333
       \]
       This is also an integer, so it is possible.

Since $a = 1$ and $b = 3$ is the only pair that satisfies all conditions, we can uniquely determine $a$ and $b$.

The final answer is $\boxed{(a, b) = (1, 3)}$