It helps to first establish the given probabilities of the events described in the question.
First, we have $\Pr[U] = \Pr[R] = 1/2$; the probability that a randomly selected customer is urban is equal to the probability that a randomly selected customer is rural, and both are equal to $1/2$.
We are also given that $\Pr[A_n \mid U] = \mu$ and $\Pr[A_n \mid R] = \lambda$; that is, the probability of a given urban customer having an accident in year $n$ is $\mu$, and the probability of a given rural customer having an accident in year $n$ is $\lambda$.
We are asked to find $\Pr[A_2 \mid A_1]$. To do this, we must use Bayes' rule: $$\Pr[X \mid Y] = \frac{\Pr[Y \mid X]\Pr[X]}{\Pr[Y]},$$ for two events $X$ and $Y$. Applying this with the choice $X = U$, $Y = A_1$, we get $$\Pr[U \mid A_1] = \frac{\Pr[A_1 \mid U]\Pr[U]}{\Pr[A_1]} = \frac{\mu/2}{\Pr[A_1]}.$$ But what is the unconditional probability $\Pr[A_1]$, that is, what is the chance that a randomly selected customer has an accident in the first year? Clearly, it relates to all four probabilities we were given--the probability of choosing an urban vs. rural customer; and the probabilities of customers of each type having an accident. The answer is to use the law of total probability: $$\Pr[X] = \Pr[X \mid Y_1]\Pr[Y_1] + \Pr[X \mid Y_2]\Pr[Y_2] + \cdots + \Pr[X \mid Y_n]\Pr[Y_n],$$ for an event $X$ and for a set of mutually disjoint events $Y_1, Y_2, \ldots, Y_n$ that satisfy $\Pr[Y_1] + \Pr[Y_2] + \cdots + \Pr[Y_n] = 1$. This is the discrete, finite version of the law. Applied to this case, we see that $$\Pr[A_1] = \Pr[A_1 \mid U]\Pr[U] + \Pr[A_1 \mid R]\Pr[R] = \frac{\mu}{2} + \frac{\lambda}{2} = \frac{\mu+\lambda}{2}.$$ Therefore, $$\Pr[U \mid A_1] = \frac{\mu}{\mu+\lambda}.$$ And similarly, $$\Pr[R \mid A_1] = \frac{\lambda}{\mu+\lambda}.$$ Now to find $\Pr[A_2 \mid A_1]$, it serves to condition $A_2$ on the customer type: $$\Pr[A_2 \mid A_1] = \Pr[A_2 \mid U]\Pr[U \mid A_1] + \Pr[A_2 \mid R]\Pr[R \mid A_1] = \frac{\mu^2+\lambda^2}{\mu+\lambda}.$$ To understand why this works, the LHS says that the customer had an accident in the first year. This customer is urban with probability $\Pr[U \mid A_1] = \mu/(\mu+\lambda)$, and if so, will have another accident next year with probability $\Pr[A_2 \mid U] = \mu$. And if this customer is rural with probability $\Pr[R \mid A_1] = \lambda/(\mu+\lambda)$, they will have another accident with probability $\Pr[A_2 \mid R] = \lambda$. This is the answer to (1).
To answer (2), independence is true if $\Pr[A_2 \mid A_1] = \Pr[A_2]$; that is, having observed $A_1$ does not change the probability of observing $A_2$. Is this true? You should argue that $\Pr[A_2] = \Pr[A_1]$: why?
For part (3), use what you found in parts (1) and (2), and the hint; now that you know the correct probabilities, the hint should make sense.
For parts (4-6), you need to extend your reasoning in the previous parts. The idea is to reason that if a randomly selected driver has accidents in years $1, 2, \ldots, n-1$, then call this event $B_{n-1} = A_1 \cap A_2 \cap \ldots \cap A_{n-1}$. Then use the law of total probability and Bayes' rule again to write $$\Pr[U \mid B_{n-1}] = \frac{\Pr[B_{n-1} \mid U]\Pr[U]}{\Pr[B_{n-1} \mid U]\Pr[U] + \Pr[B_{n-1} \mid R]\Pr[R]}$$ where I have combined the two formulas into a single expression. The only difference now is that you need to know the probabilities of a driver of each type having $n-1$ consecutive accidents, rather than just a single accident.
Then repeat the above for $\Pr[R \mid B_{n-1}]$, which should be easy to see in advance; finally, follow the method at the end of part (1) to reason that $$\Pr[A_n \mid B_{n-1}] = \ldots?$$ I'm leaving this to you.
As an instructive exercise, once you finish this entire question, go back and redo everything in the generalized case where $\Pr[U] = p$, $\Pr[R] = q = 1-p$, where $0 \le p, q \le 1$.
Best Answer
This problem concerns Bayes Theorem.
The total probability that a random person has an accident in the given year is $$.8\times .01+.2\times .04=0.016$$
The portion of that which is explained by careful drivers is $.8\times .01=.08$. Thus the answer is $$\frac {.8\times .01}{.8\times .01+.2\times .04}=.5$$
Thus it is equally likely that your accident victim was careful or careless.
Speaking informally, that's because, while it's four times more likely that a randomly selected driver is careful, it is one fourth as likely that a careful driver will have an accident, and the two factors of four cancel.