Homework - Genetics linkage problem
I agree with your reasoning! If the two genes are perfectly linked, then the answer is $\frac{1}{8}$. If they are perfectly unlinked, then the answer is $\frac{1}{16}$. The answer therefore must be somewhere between $\frac{1}{8}$ and $\frac{1}{16}$.  Good job!

Of course the answer e) is pretty silly as any other answer ('a','b','c' and 'd') would yield to 'e' being necessarily true. So, if you have to choose only one, I would just go with 'd' and highlight that the answer 'e' is silly.


  I still don’t understand how if the two are completely linked, the answer would be 1/8.


Let's consider that at both loci, we have two alleles A and B. Here is a representation of the first F1 individual

-------A----A-----
-------B----B-----

Here is a representation of the second F1 individual

-------A----A-----
-------B----B-----

Of course, they are the same as the parental lines are pure breed. Now, if the two loci are perfectly linked, then the offspring can have the following possible genotypes

-------A----A-----
-------A----A-----,

-------B----B-----
-------A----A-----,

-------A----A-----
-------B----B-----

or

-------B----B-----
-------B----B-----

, but the offspring cannot be 

-------B----A-----
-------B----B-----

or an other combination requiring recombination between the two loci. Therefore the probability to be heterozygote for the F2 at two perfectly linked loci is the same as the probability to be heterozygote at one of these two loci (if we assume the mutation rate is negligible).


  Why can't parents be: parent1: -------B----A----- -------A----B----- parent2: -------A----A----- -------B----B-----[?]


Because parental lines are pure lines. When we talk about an F1 generation, we talk about the hybrid between two pure line. By pure line, we mean that we got rid of all genetic variance. So, one parental line is fixed for the A alleles and the other line is fixed for the B alleles.  In other words here are the genotypes of the pure parental lines. 

-------A----A-----
-------A----A-----

and

-------B----B-----
-------B----B-----

One parent can therefore only give 

-------A----A-----

while the other one can only give 

-------B----B-----

so, the F1 generation is therefore necessarily heterozygote at all loci.