Concept: Energy stored in a spring under constant force. Formula: \(PE = \frac{F^2}{2K}\). When the same force \(F\) is applied, potential energy is inversely proportional to the spring constant (\(PE \propto 1/K\)). Given \(K_A = 2K_B\). The ratio \(\frac{PE_B}{PE_A} = \frac{K_A}{K_B}\). Substituting \(K_A = 2K_B\), we get \(\frac{PE_B}{E} = \frac{2K_B}{K_B} = 2\). Therefore, \(PE_B = 2E\).

Concept: Work done to stretch a spring under a constant force. Formula: \(W = \frac{F^2}{2K}\). When the same force \(F\) (due to suspended weights) is applied, the work done is inversely proportional to the spring constant (\(W \propto 1/K\)). Given \(K_A = 2K_B\). The ratio of work done is \(\frac{W_A}{W_B} = \frac{1/K_A}{1/K_B} = \frac{K_B}{K_A}\). Substituting \(K_A = 2K_B\), we get \(\frac{W_A}{W_B} = \frac{K_B}{2K_B} = \frac{1}{2}\). So the ratio is \(1:2\).

Concept: Hooke's Law. Formula: \(F = k(L - L_0)\), where \(L_0\) is original length. We have: (1) \(4 = k(a - L_0)\), (2) \(5 = k(b - L_0)\). From (1) and (2), we find \(L_0 = 5a - 4b\) and \(k = \frac{1}{b - a}\). For \(F=9\text{ N}\), \(9 = k(L_3 - L_0)\). Substitute \(k\) and \(L_0\): \(9 = \frac{1}{b - a}(L_3 - (5a - 4b))\). Solving for \(L_3\), we get \(L_3 = 9(b - a) + 5a - 4b = 9b - 9a + 5a - 4b = 5b - 4a\).