Hello Melissa,
Hooke's law is given as:
(1) $$F=-kx$$
where $x$ is the displacement of the spring from equilibrium and $k$ is a positive constant called the force constant of the spring. Solving (1) for $k$, we may write:
$$k=-\frac{F}{x}$$
We see that $k$ will have units of force per length, and in the case of this problem, this will be pounds per inch. I will also give the results in pounds per feet.
a) If we consider a compression of the spring to be a negative displacement, then using the fact that a 150 lb. person compresses the spring 1/14 in., we the find:
$$k=\frac{150}{1/14}\,\frac{\text{lb}}{\text{in}}=2100\,\frac{\text{lb}}{\text{in}}=25200\,\frac{\text{lb}}{\text{ft}}$$
b) Using (1), and the result from part a), we find:
$$F=-\left(2100\,\frac{\text{lb}}{\text{in}} \right)\left(-\frac{1}{8}\text{ in} \right)=262.5\text{ lb}$$
c) To compute the work done in compressing the spring 1/8 in, we may use:
$$W=\int_{-s}^0 F_s\,dx=-k\int_{-s}^0 x\,dx=-\frac{k}{2}\left[x^2 \right]_{-s}^0=\frac{k}{2}s^2$$
With $$s=\frac{1}{8}\text{ in}$$ we find:
$$W=\frac{2100\,\frac{\text{lb}}{\text{in}}}{2}\left(\frac{1}{8}\text{ in} \right)^2=\frac{525}{32}\text{ in}\cdot\text{lb}=\frac{175}{128}\text{ ft}\cdot\text{lb}$$
