In this problem, a spring of spring constant $k = 85~mathrm{N/m}$ is stretched to have a potential energy of $PE_text{spring} = 0.22~mathrm{J}$. We calculate how long the spring is stretched.

From the definition of elastic potential energy, we have

$$
begin{align*}
PE_text{spring} &= frac{1}{2}kx^{2} \
x^{2} &= frac{2PE_text{spring}}{k} \
implies x &= sqrt{frac{2PE_text{spring}}{k}} \
&= sqrt{frac{2 left( 0.22~mathrm{J} right)}{85~mathrm{N/m}}} \
&= 7.19477~mathrm{cm} \
x &= boxed{ 7.2~mathrm{cm} }
end{align*}
$$

$$
x = 7.2~mathrm{cm}
$$

In this problem, a hockey stick stores a potential energy of $PE_text{spring} = 4.2~mathrm{J}$ when bent $x = 3.1~mathrm{cm} = 3.1 times 10^{-2}~mathrm{m}$. Treating it like a spring, we calculate the spring constant.

From the definition of elastic potential energy, we have

$$
begin{align*}
PE_text{spring} &= frac{1}{2}kx^{2} \
k &= frac{2PE_text{spring}}{x^{2}} \
&= frac{2 left( 4.2~mathrm{J} right)}{left( 3.1 times 10^{-2}~mathrm{m} right)^{2}} \
&= 8740.89490~mathrm{N/m} \
k &= boxed{ 8700~mathrm{N/m} }
end{align*}
$$

$$
k = 8700~mathrm{N/m}
$$