For $n\in \{0,1,2\}$ Let $E[n]$ denote the answer given that you are starting from a streak of $n$ consecutive rolls. The answer you want is $E=E[0]$, though you are never in state $0$ except at the start.
We note $$E[2]=\frac 16\times 1+\frac 56\times \left(E[1]+1\right)$$
$$E[1]=\frac 16\times \left(E[2]+1\right)+\frac 56\times \left(E[1]+1\right)$$
$$E=E[0]=E[1]+1$$
this system is easily solved and, barring error (always possible), yields $$\boxed {E=43}$$
Jess picks $3$ days. We want the probability that Lindsay picks a day that Jess has picked. It's easier to compute the complement. The probability that Lindsay picks both days from the two that Jess hasn't picked is $${{4\choose2}\over{7\choose2}}={{4\cdot3}\over{7\cdot6}}=\frac27$$ so the probability we seek is $$1-\frac27=\frac57$$
EDIT
When I woke up this morning, I realized how you were attempting to do the problem, or at least I think I do. You want to say, for any five days of the week, compute the probability that Jess and Lindsay pick those five days, and then multiply by the number of ways to choose five days. That will work, but you've computed the probability incorrectly.
Suppose the five days have been chosen. Jess can choose three of these days in ${5\choose3}$ ways and he can choose three days of the week in ${7\choose3}$ ways, so the probability of success is $${{5\choose3}\over{7\choose3}}$$ Then Lindsay must choose the remaining two days, and he has ${7\choose2}$ choices, so the probability that he chooses the remaining two days is $${1\over{7\choose2}}$$ Now, we must add these probabilities up for every possible choice of five days and there are ${7\choose5}={7\choose2}$ such choices. Altogether, we have a probability of $$
{{5\choose3}\over{7\choose3}}{1\over{7\choose2}}{7\choose2}=
{{5\choose3}\over{7\choose3}}=
{{5\cdot4\cdot3}\over{7\cdot6\cdot5}}={2\over7}$$
I think that the basic flaw in your approach is trying to do the problem too quickly. It seems to me that you are writing down calculations without stopping to think what they mean, or why they should be correct. This is very tempting, I know, but you should resist the temptation. Try to get into the habit of writing down, in words, what you are doing, and why it is correct. If you don't, you may find that when you go back and look at the calculation later, you may find that you can't understand it, even if it is correct, or that you can't figure out why you thought it was correct, if it was wrong. This has happened to me many times.
Also, of course, if you don't do this, it is harder to communicate with others. Look how long it's taken me to figure out the gist of your calculations. Somebody smarter than I would have seen it it a lot sooner no doubt, and I may have been sleepy yesterday, but even so, it shouldn't have taken so long.
I hopes this helps.
Best Answer
In your stated problem, "counting" is counting weeks where any tie was chosen twice.
After you do your trials, rolling the dice for X number of week sequences, you will end up with a number of weeks that a tie was chosen twice / number of week sequences you ran for your Monte Carlo series, and this value will be your estimated probability that in a future week a tie will be chosen twice.
The idea of Monte Carlo simulations is that when you have some sort of random set of events that are too large (or perhaps infinite) to exhaustively calculate all finite probabilities, so by running a number of random tests you can start to approximate the probability. The larger your number of tests, the better the approximation is.
Your particular assigned problem is pretty straightforward to calculate exhaustively, though to list out the 3125 different permutations would be a bit tedious. What the assignment is trying to show you is how many "weeks" you have to simulate to start to approach the final probability.