To solve this problem, we need to consider the forces acting on the barge due to the tensions \(T_1\) and \(T_2\). The resultant force along the direction of travel of the barge is given as 5000 N.

Here’s the approach to solve the problem step-by-step:

1. Let \(\theta_1 = 30^\circ\) be the angle \(T_1\) makes with the direction of travel, and \(\theta_2 = 45^\circ\) be the angle \(T_2\) makes with the direction of travel.
2. The horizontal components of \(T_1\) and \(T_2\) can be expressed using trigonometry:
   • \(T_{1x} = T_1 \cdot \cos(30^\circ)\)
   • \(T_{2x} = T_2 \cdot \cos(45^\circ)\)
3. According to the problem, the sum of these horizontal components results in a force of 5000 N:
   • \(T_1 \cdot \cos(30^\circ) + T_2 \cdot \cos(45^\circ) = 5000\)
4. Using the cosine values, the equation becomes:
   • \(T_1 \cdot \frac{\sqrt{3}}{2} + T_2 \cdot \frac{\sqrt{2}}{2} = 5000\)
5. Along the vertical direction, the barge is in equilibrium since there is no vertical movement, thus:
   • \(T_1 \cdot \sin(30^\circ) = T_2 \cdot \sin(45^\circ)\)
6. Using the sine values, we have:
   • \(T_1 \cdot \frac{1}{2} = T_2 \cdot \frac{\sqrt{2}}{2}\)
7. Solving these equations simultaneously:
   • From \(T_1 \cdot \frac{\sqrt{3}}{2} + T_2 \cdot \frac{\sqrt{2}}{2} = 5000\), and
   • \(T_1 \cdot \frac{1}{2} = T_2 \cdot \frac{\sqrt{2}}{2}\), leading to \(T_1 = T_2 \sqrt{2}\).
8. Substituting \(T_1 = T_2 \sqrt{2}\) in the first equation:
   • \((T_2 \sqrt{2}) \cdot \frac{\sqrt{3}}{2} + T_2 \cdot \frac{\sqrt{2}}{2} = 5000\)
   • After simplifying, we get \(T_2 = 2588\) N and \(T_1 = 3660\) N.

Thus, the tensions are 3660 N and 2588 N respectively.

The correct answer is: 3660 and 2588.