A Poisson random variable is frequently used to model the probability of observing a count or frequency; i.e., "how many times did something happen?" It is a discrete random variable--we don't say "we observed exactly 3.4 car accidents in three days," or "there were 4.7 bad apples in a bucket."
An exponential random variable is used to model the probability of a failure time, or event time; i.e., "how long did it take for something to happen?" It is a continuous random variable because we can measure time in very fine increments that do not have to be integers; e.g., "the phone failed after 3.26 years of use."
When a Poisson process is used to model the occurrence of an event of interest, it happens that the frequency distribution is Poisson, but the interarrival times are exponentially distributed; that is, the random number of events that occur in a given period of time follows a Poisson distribution, but the random time that elapses between one observed event and the next is exponentially distributed.
Therefore, the time $T$ that elapses until you observe the first factory accident is exponentially distributed with rate parameter $\lambda = 2$ accidents per week. The survival function of $T$ is $S_T(t) = e^{-\lambda t}$, so we want $$\Pr[T > 2] = S_T(2) = e^{-2(2)} = e^{-4}.$$ The other part of your question is addressed similarly, except you need to convert $2$ days into an equivalent (fractional) number of weeks.
Now that you have indicated your work, I can write out a solution to the problem. Recall that if $A$ and $B$ are events, with $\Pr(B)\ne 0$, then the conditional probability $\Pr(A|B)$ of $A$ given $B$ is given by
$$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}.$$
Let $B$ be the event "at least one accident" and let $A$ be the event "more than two accidents."
We have $\Pr(B)=1-e^{-\lambda}$, where $\lambda=1.5$.
The event $A\cap B$ occurs if there are more than $2$ accidents. Its probability is $1$ minus the probability of $2$ or fewer accidents. It follows that
$$\Pr(A\cap B)=1-e^{-\lambda}\left(1+\frac{\lambda}{1!}+\frac{\lambda^2}{2!}\right).$$
Finally, divide, as in Equation (1).
For b), use the fact that the number of accidents in time $t$ has Poisson distribution with parameter $\lambda t$, where $\lambda$ is the mean number of accidents per unit time, here $1.5$. The probability of at least one accident is $1-e^{-\lambda t}$. Set this equal to $0.95$.
We get $e^{-\lambda t}=0.05$. Solve for $t$, by taking the natural logarithm of both sides.
Remark: I feel a little guilty about asserting without proof the fact about the distribution of the number of accidents in $t$ days. For our problem, we do not need the full fact. Suppose that the numbers of accidents from day to day are independent random variables. Then, since the probability of no accidents in $1$ day is $e^{-\lambda}$, the probability of no accidents in $n$ days, where $n$ is a positive integer is $(e^{-\lambda})^n$, which is $e^{-\lambda n}$.
It is reasonable to believe (and true) that even if $t$ is not an integer, the probability of no accidents in $t$ days is $(e^{-\lambda})^t$.
Best Answer
There are seven days and seven accidents (which are distinguishabla from one another), so the number of possible outcomes is $7^7$. However, only seven of them fulfil our criteria: all accidents on Monday, all on Tuesday, ..., all on Saturday. So the answer is $1/7^6$.