To solve for the tension \(T\) and the mass \(m\), we resolve forces horizontally and vertically.

Horizontally, the equilibrium condition gives:

\(T \sin 30^{\circ} + T \sin 40^{\circ} - 2 = 0\)

Vertically, the equilibrium condition gives:

\(T \cos 30^{\circ} - T \cos 40^{\circ} - mg = 0\)

Solving the horizontal equation for \(T\):

\(T (\sin 30^{\circ} + \sin 40^{\circ}) = 2\)

\(T = \frac{2}{\sin 30^{\circ} + \sin 40^{\circ}}\)

Substituting the values of \(\sin 30^{\circ} = 0.5\) and \(\sin 40^{\circ} \approx 0.6428\):

\(T = \frac{2}{0.5 + 0.6428} \approx 1.75\)

Substitute \(T = 1.75\) into the vertical equation:

\(1.75 \cos 30^{\circ} - 1.75 \cos 40^{\circ} = mg\)

\(mg = 1.75 (\cos 30^{\circ} - \cos 40^{\circ})\)

Using \(\cos 30^{\circ} \approx 0.866\) and \(\cos 40^{\circ} \approx 0.766\):

\(mg = 1.75 (0.866 - 0.766)\)

\(mg = 1.75 \times 0.1 = 0.175\)

\(m = \frac{0.175}{g}\)

Assuming \(g = 9.8\):

\(m \approx \frac{0.175}{9.8} \approx 0.0175\)