Two Springs in Parallel Two identical springs in parallel are supporting a mass. One spring is twice as stiff. Which spring experiences more force?
I want to say that the stiffer spring experiences more force using Hooke's law but I am unsure.
 A: If this is an interview question then it is probably intentionally underspecified to see what questions you ask.
One of the pieces of information that is missing is how the attachment points between the springs and the mass are arranged. We don’t know whether the attachment points are symmetrically placed or not. But if we assume they are symmetrically placed then the forces exerted on the mass by the two springs must be the same (otherwise there will be a turning moment on the mass and it will not be in equilibrium). However, the extensions of the two springs will not be the same - the stiffer spring will extend by a smaller amount.
A: Assume that a mass of $m$ kg is hung against gravity by these two springs (in parallel)
For first spring, spring constant = $k$
For second spring (stiffer), spring constant = $2k$
Now in equilibrium position net force on the mass is zero.
So,
$$F_{sp_1} + F_{sp_2} = mg$$
$$kx + 2kx = mg$$
Note that both the spring will elongate with same length from their natural length position
Now it is obvious from the second equation that the force exerted by the second spring (stiffer) is more.
Hope this helps.
A: Spring constant, 
For parallel: $$k_{tot}=\sum_i k_i$$
For series
$$k_{tot}=(\sum_i \frac{1}{k_i})^{-1}$$
Since you have two spring attached parallel. So spring constant for first one will be $k$ for second one $$k_{2nd}=k+k=2k$$ If spring constant for second one is different then you can write $k_{2nd}=k+k_2$
$$F_{2nd}=-k_{2nd}x$$

Which spring will experience more force?

That depends on value of $k$. First one will experience more force.
$$F_1=-kx>-2kx=F_2$$
for the case $-1>-2$
