(a) To find the tensions in the strings, resolve the forces horizontally and vertically. Let the tensions in strings AC and BC be T_A and T_B respectively.

Resolving horizontally: \(T_A \times 0.6 - T_B \times 0.8 = 20\)

Resolving vertically: \(T_A \times 0.8 + T_B \times 0.6 = 10g\)

Solving these equations simultaneously:

From the horizontal resolution: \(T_A = \frac{0.8}{0.6} T_B + \frac{20}{0.6}\)

Substitute into the vertical resolution:

\(\left( \frac{0.8}{0.6} T_B + \frac{20}{0.6} \right) \times 0.8 + T_B \times 0.6 = 10g\)

Simplifying gives: \(0.87 T_A = 20 \rightarrow T_A = 76 \, \text{N}\)

Substitute back to find \(T_B\): \(T_B = 68 \, \text{N}\)

(b) For the greatest value of F, consider when \(T_B = 0\):

From the vertical resolution: \(T_A \times 0.6 = 10g \rightarrow T_A = \frac{500}{3}\)

From the horizontal resolution: \(T_A \times 0.8 = F\)

Substitute \(T_A = \frac{500}{3}\):

\(F = \frac{400}{3} \approx 133 \, \text{N}\)