Suppose Thelma asks her mother on the first day.
If she gets permission, then on the second day she should definitely not ask her mother again, but ask dad instead. If he consents, then Thelma will be going out on two consecutive days, and otherwise she won't (regardless of the result of the third day).
If mother does not give permission on the first day, then Thelma should ask her again on the second day because she is then certain to go out. On the third day Thelma can't ask her again, so it once again depends on dad whether she goes out two days in a row.
Regardless of the outcome when Thelma asks mum first, the probability of going out on two consecutive days is the probability of dad consenting once.
By symmetry, if Thelma asks dad first, the probability of going out for two consecutive days is the probability of mum consenting once.
Since dad is more lenient, Thelma should ask mum first.
To recap: Ask mum on the first day. If yes, ask dad the next day, if no, ask mum again on the second day and then dad on the third day.
I originally misread the question, thinking that the rule was that if you ask one parent on two consecutive days, they won't say yes twice. This is a pretty interesting question too, so I'll keep the answer for that question below:
Clearly she must alternate which parent she asks, because if she asks the same parent twice in a row it definitely won't happen for that pair of days. So she must decide between the order mum-dad-mum or dad-mum-dad.
Before going into the maths, Thelma might intuitively think the latter is better, because the dad is more lenient and so she is likely to go out more days.
But that does not necessarily mean she is more likely to go out on two consecutive days. In fact, the first combination is better, because she must go out on the middle day, so that is best asked to the more lenient parent, and then she has two chances of getting the strict parent to consent on one of the other days.
Here is the maths:
Let $m$ be the probability that mother gives permission when asked, and $d$ the same for her father. We are given that $m<d$.
To go out two consecutive days when asking dad-mum-dad, she needs mum to consent, and dad not to refuse twice. This probability is $m(1-(1-d)^2) = dm(2-d)$. If we ask mum-dad-mum we get the same but with the letters swapped, i.e. $md(2-m)$. Given that $m<d$, the latter probability is largest.