To solve for the tensions in the strings, we resolve the forces acting on P both horizontally and vertically.

1. **Horizontal Resolution:**

The horizontal component of the tension in the string attached to A is \(T_1 \cos 36.9^ ext{o}\) and for the string attached to B is \(T_2 \cos 16.3^ ext{o}\). The horizontal force is 10 N.

Thus, the equation is:

\(T_1 \cos 36.9^ ext{o} + T_2 \cos 16.3^ ext{o} = 10\)

2. **Vertical Resolution:**

The vertical component of the tension in the string attached to A is \(T_1 \sin 36.9^ ext{o}\) and for the string attached to B is \(T_2 \sin 16.3^ ext{o}\). The weight of P is \(0.7g\) N.

Thus, the equation is:

\(T_1 \sin 36.9^ ext{o} - T_2 \sin 16.3^ ext{o} = 0.7g\)

3. **Solving the Equations:**

Using the given values:

\(0.8T_1 + 0.96T_2 = 10\)

\(0.6T_1 - 0.28T_2 = 0.7g\)

Solving these simultaneous equations gives:

\(T_1 = 11.9\) N and \(T_2 = 0.5\) N.