Let the probability that it actually rains be $R$. Our prior information about $R$ is based on the "past few years" and it tells us that it rains, on average, 5 days a year (let's say three non-leap years just to finesse the problem of leap days) so we know that:
$$
P(R) = \frac{5}{365} = \frac{1}{73} \approx .0137
$$
Now let's talk about the weather forecaster's accuracy (or lack thereof). Let $P$ represent the probability that Mr. Sunshine accurately predicts rain. We know that his true positive rate (accurate prediction of presence of event) is 90%, so we know $P(P|R)$, the probability that his prediction of rain occurs when (given) it actually rains. We also know his false positive rate, the probability that the test (his prediction) indicates true when the phenomenon (rain) is false, $P(P|\overline{R})$ is 10%.
Lastly, poor Marie is in possession of a $P$, a prediction of rain. She wants to know $P(R|P)$, the probability that it will rain given the weather report. Now we can use Bayes Theorem:
$$
\begin{align}
P(R|P) &= \frac{P(R \cap P)}{P(P)}\\
P(R \cap P) &= P(P \cap R) = P(P|R)P(R)\\
P(P) &= P(P \cap R) + P(P \cap \overline{R})\\
P(P \cap \overline{R}) &= P(P|\overline{R})P(\overline{R})
\end{align}
$$
So, fully spelled out in terms of the data we have, we need:
$$
P(R|P) = \frac{P(P|R)P(R)}{P(P|R)P(R) + P(P|\overline{R})P(\overline{R})}
$$
Which when we substitute values becomes
$$
\begin{align}
P(R|P) &=\frac{\frac{9}{10}\times \frac{1}{73}}{\left(\frac{9}{10}\times\frac{1}{73}\right)+\left(\frac{1}{10}\times \frac{72}{73}\right)}\\
&=\frac{\frac{9}{730}}{\frac{9}{730} + \frac{72}{730}}\\
&=\frac{9}{81} = \frac{1}{9} \approx 11.1\%
\end{align}
$$
Denoting the current day as index $i$ (so that yesterday is $i-1$ and the day before yesterday is $i-2$), the previous state comprises two elements, $State_{n-1}=\{S_{i-1},S_{i-2}\}$, where $S_k$ is the state of the weather for day $k$. Thus the current state is $State_{n}=\{S_{i},S_{i-1}\}$
Lettting $R$ be the occurrence of rain on a particular day, so that $\overline{R}$ is when no rain occurs for that day, there are $4$ possible values of the previous state $State_{n-1}$
$$\{\overline{R},\overline{R}\}, \{\overline{R},R\},\{R,\overline{R}\},\{R,R\}$$
For each of these $4$ states, there are only two possible next states:-
$$\begin{align}
State_n=\{\overline{R},\overline{R}\}\text { or }State_n=\{R,\overline{R}\}\text{ given }State_{n-1}=\{\overline{R},\overline{R}\}
\\
State_n=\{\overline{R},\overline{R}\}\text { or }State_n=\{R,\overline{R}\}\text{ given }State_{n-1}=\{\overline{R},R\}
\\
State_n=\{\overline{R},R\}\text { or }State_n=\{R,R\}\text{ given }State_{n-1}=\{R,\overline{R}\}
\\
State_n=\{\overline{R},R\}\text { or }State_n=\{R,R\}\text{ given }State_{n-1}=\{R,R\}
\end{align}$$
With the information given in the question, and based on the constraints of what the next state is based on the current state, you can figure out the structure of the transition matrix.
For example, the information Specifically, suppose that if it has rained for the past two days, then it will rain tomorrow with probability $0.7$, corresponds to
$$P(State_{n+1}=\{\overline{R},\overline{R}\}|State_{n}=\{\overline{R},\overline{R}\})=0.7\\\Rightarrow P(State_{n+1}=\{R,\overline{R}\}|State_{n}=\{\overline{R},\overline{R}\})=1-0.7=0.3$$
and if it rained today but not yesterday, then it will rain tomorrow with probability 0.5 corresponds to
$$P(State_{n+1}=\{\overline{R},\overline{R}\}|State_{n}=\{\overline{R},R\})=0.5\\\Rightarrow P(State_{n+1}=\{R,\overline{R}\}|State_{n}=\{\overline{R},R\})=1-0.5=0.5$$
Best Answer
Let $F$ represent the event that he forecasted rain will occur. Let $R$ represent that rain actually occurs.
We are told the following information:
$Pr(F\mid R)=0.9$ (if it rained, the forecaster predicted it correctly 90% of the time)
$Pr(F\mid R^c)=0.1$ (if it didn't rain, the forecaster incorrectly predicted it should have rained 10% of the time)
We are also likely intended to assume that $Pr(R)=\dfrac{5}{365}$ as per the phrase "in recent years it has rained only 5 days each year" or perhaps $Pr(R)=\dfrac{5}{365.25}$ or some other similar number according to how much you wish to account for leap year. For simplicity's sake, I would suggest using $\dfrac{5}{365}$.
We are tasked with calculating $Pr(R\mid F)$.
Now... with this information we can apply Bayes' theorem:
$$Pr(R\mid F)=\dfrac{Pr(F\mid R)Pr(R)}{Pr(F)}$$
The terms in the numerator were given directly in the problem. The term in the denominator can be calculated using the law of total probability and conditional probability:
$Pr(F)=Pr(F\cap R)+Pr(F\cap R^c)=Pr(R)Pr(F\mid R)+Pr(R^c)Pr(F\mid R^c)$
Plug in the appropriate numbers and reach a final conclusion.