To solve for the tensions in the strings, we resolve the forces horizontally and vertically.

Let the tensions in the strings be \(T_1\) and \(T_2\).

Resolving horizontally:

\(T_1 \cos 60^{\circ} = T_2 \cos 45^{\circ}\)

Resolving vertically:

\(T_1 \sin 60^{\circ} + T_2 \sin 45^{\circ} = 8g\)

Substituting \(g = 9.8 \text{ m/s}^2\), we have:

\(T_1 \cos 60^{\circ} = T_2 \cos 45^{\circ}\)

\(T_1 \sin 60^{\circ} + T_2 \sin 45^{\circ} = 8 \times 9.8\)

Solving these equations simultaneously gives:

\(T_1 = 58.6 \text{ N}\)

\(T_2 = 41.4 \text{ N}\)

An alternative method using Lami's Theorem:

\(\frac{T_1}{\sin 135^{\circ}} = \frac{T_2}{\sin 150^{\circ}} = \frac{80}{\sin 75^{\circ}}\)

Solving for \(T_1\) and \(T_2\) gives the same results:

\(T_1 = 58.6 \text{ N}\)

\(T_2 = 41.4 \text{ N}\)