Not sure if this properly captures the dynamics of mummy-chasing, but a reasonable assumption is that curses get tripped, with concomitant mummy-chasing, following a homogeneous Poisson point process, which is characterized by two assumptions:
(a) Over any time interval $[a,b]$ the number $N(a,b)$ of times the mummy arrives is a Poisson random variable with mean $\lambda(b-a)$, where $\lambda>0$ is some constant that represents the rate or intensity of arrivals. ($\lambda$ is also interpretable as the expected number of chases per unit time.) Note that the mummy can arrive more than once in a given time period, but this has low probability.
(b) The number of mummy arrivals in disjoint time intervals are independent random variables.
If we measure time in days, then in a given day the number of mummy arrivals would be Poisson($\lambda$) by assumption (a). So the probability of no chase today would be $e^{-\lambda}$, hence the probability of being chased today is $\fbox{$1-e^{-\lambda}$}$.
Your mission is to find this probability (i.e., find $\lambda$) under various assumptions about the distribution of trip duration, knowing that chases occur on 10% of trips. In your simplest case, where every trip lasts 14 days, use the independence assumption (b) to conclude the probability of no chase during the trip is $(e^{-\lambda})^{14}$. Equate this to $0.9$, solve for $\lambda$ and plug this into the daily probability of chase. This is exactly your initial argument, with $x:=e^{-\lambda}$.
In the general case the trip duration $T$ is a random variable with some density $f$. When the trip duration $T$ has value $t$, the probability of no chase during the entire trip is
$\fbox{$e^{-\lambda t}$}$, so averaging over all trip durations we obtain
$$P(\text{no chase})=\int P(\text{no chase}\mid T=t)f(t)\,dt=\int e^{-\lambda t}f(t)\,dt.$$
I think you are to assume that $T$ has a normal distribution with mean $\mu$ and standard deviation $\sigma$. If so the RHS of the above evaluates to $\exp(-\mu\lambda + \frac12\sigma^2\lambda^2)$. Since chases occur on just 10% of trips, this gives the equation:
$$\bbox[5px,border:1px solid black]{0.9=P(\text{no chase})=\exp(-\mu\lambda + \frac12\sigma^2\lambda^2).}$$
This is the main equation you need to solve for $\lambda$. To do so you need to identify $\mu$ and $\sigma$.
Your assumptions (1) and (2) give you enough information to identify $\mu$ and $\sigma$: (1) is saying $\mu=14$, and (2) says
$$P(|T-\mu|<2)=0.9.$$ Since the LHS of the above equation is the same as $P(|Z|<\frac2\sigma)$ for $Z$ standard normal, and since $P(|Z|<1.645)=0.9$, you deduce $\frac2\sigma=1.645$ and solve for $\sigma$.
You can handle the other scenarios the same way. Your scenario (3a) can be interpreted as $P(|T-\mu|<0.5)=0.4$ (so that 40% of trips end within a specific 24-hour period -- note the fact that this day is Saturday is immaterial). Similarly your scenario (3b) is saying $\sigma=30/24$. Given either of these assumptions you can solve for $\lambda$ under the assumption $\mu=14$.
(If you're not comfortable with the Poisson process, you can repeat the above analysis writing $x:=e^{-\lambda}$. It's enough to assume that the probability you are not chased during a trip of duration $T=t$ is $x^t$. The above argument will carry through.)
