Pls help me with 12 and 13 question

Dear student,
The solution/s to your 13th question has/have been provided below:
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$$\begin{gathered} Stored potential energy = \frac{1}{2} K{l}^2 , for an extension l. \\ Work done = change in stored potential energy (by conservation of energy) \\ \frac{1}{2} K{(l+l)}^2 - \frac{1}{2} K{l}^2 \\ = \frac{1}{2} K{\left(2l\right)}^2 - \frac{1}{2} K{l}^2 =\frac{1}{2} K{l}^2 (4-1) =\frac{3}{2} K{l}^2 \end{gathered}$$


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