Answer: no.

Under all transformations $(a, b, c, d) \rightarrow\left(a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime}\right)$ allowed in the problem we have $|a d-b c|=\left|a^{\prime} d^{\prime}-b^{\prime} c^{\prime}\right|$, but $|1 \cdot 4-2 \cdot 3|=2 \neq 1=|3 \cdot 7-4 \cdot 5|$.

Remark. The transformations allowed in the problem are in fact the elementary transformations of the determinant

$$\left|\begin{array}{ll} a & b \\ c & d \end{array}\right|$$

and the invariant $|a d-b c|$ is the absolute value of the determinant which is preserved under these transformations.

1. Define the invariant: For a quadruple $(p, q, r, s)$, let $N$ denote $|ps - qr|$. We will show that $N$ is invariant under the given transformations.

2. Check invariance under each transformation:
- For the transformation $(a, b, c, d) \to (c, d, a, b)$:
$$N = |ps - qr| = |ad - bc|$$
After transformation, we have:
$$N' = |cb - da| = |ad - bc| = N$$
- For the transformation $(a, b, c, d) \to (b, a, d, c)$:
$$N = |ps - qr| = |ad - bc|$$
After transformation, we have:
$$N' = |ba - dc| = |ad - bc| = N$$
- For the transformation $(a, b, c, d) \to (a + nc, b + nd, c, d)$:
$$N = |ps - qr| = |ad - bc|$$
After transformation, we have:
$$N' = |(a + nc)d - (b + nd)c| = |ad + ncd - bc - ndc| = |ad - bc| = N$$
- For the transformation $(a, b, c, d) \to (a + nb, b, c + nd, d)$:
$$N = |ps - qr| = |ad - bc|$$
After transformation, we have:
$$N' = |(a + nb)d - b(c + nd)| = |ad + nbd - bc - bnd| = |ad - bc| = N$$

3. **Calculate $N$ for the initial and target quadruples**:
- For the initial quadruple $(1, 2, 3, 4)$:
$$N = |1 \cdot 4 - 2 \cdot 3| = |4 - 6| = 2$$
- For the target quadruple $(3, 4, 5, 7)$:
$$N = |3 \cdot 7 - 4 \cdot 5| = |21 - 20| = 1$$

4. Compare the invariants: Since $N$ is invariant under the given transformations, and the initial $N$ is 2 while the target $N$ is 1, it is impossible to transform $(1, 2, 3, 4)$ into $(3, 4, 5, 7)$ using the given transformations.

$\blacksquare$

The final answer is False