(a) The force $$mg$$ is the tension in each of the springs. The bottom of the upper (first) spring moves down by distance $$x_{1} = |F|/k_{1} = mg/k_{1}$$. The top of the second spring moves down
by this distance, and the second spring also stretches by $$x_{2} = mg/k_{2}$$. The bottom of the lower spring then moves down by distance
$$x_{total}=x_{1}+x_{2}=\dfrac{mg}{k_{1}}+\dfrac{mg}{k_{2}}=mg\left ( \dfrac{1}{k_{1}}+\dfrac{1}{k_{2}} \right )$$
(b) From the last equation we have
$$mg=\dfrac{x_{1}+x_{2}}{\dfrac{1}{k_{1}}+\dfrac{1}{k_{2}}}$$
This is of the form
$$|F|=\left ( \dfrac{1}{1/k_{1}+1/k_{2}} \right )(x_{1}+x_{2})$$
The downward displacement is opposite in direction to the upward force the springs exert on the load, so we may write $$F = –k_{eff} x_{total}$$, with the effective spring constant for the pair of springs given by
$$k_{eff}=\dfrac{1}{1/k_{1}+1/k_{2}}$$