Answer
(a) $k = \frac{k_1~k_2}{k_1+k_2}$
(b) The elastic potential energy stored in the spring is $0.15~J$
Work Step by Step
(a) Let's consider the two springs attached in series. When a force of $F$ is exerted on the springs, both springs pull with a force of $F$. Suppose the total stretch distance of the system is $x$.
$F = k_1~x_1 = k_2~x_2$
$k_1~x_1 = k_2~(x-x_1)$
$x = \frac{(k_1+k_2)~x_1}{k_2}$
We can find the effective spring constant $k$ of the combination:
$k = \frac{F}{x} = \frac{k_1~x_1}{\frac{(k_1+k_2)~x_1}{k_2}} = \frac{k_1~k_2}{k_1+k_2}$
(b) We can find the value of $k$:
$k = \frac{k_1~k_2}{k_1+k_2}$
$k = \frac{(500~N/m)(300~N/m)}{500~N/m+300~N/m}$
$k = 187.5~N/m$
We can find the elastic potential energy:
$U_s = \frac{1}{2}kx^2$
$U_s = \frac{1}{2}(187.5~N/m)(0.040~m)^2$
$U_s = 0.15~J$
The elastic potential energy stored in the spring is $0.15~J$
