Probability it rains

The point in Probabilities is the following: Denote $W$ the event that it rains on Wednesday, ${\bar W}$ the event that it doesn't. Likewise with $T$ and $\bar T$ for Thursday.

Then $P(W) =0.4$.

By marginalization, you have $0.3 = P(T) = P(T|W) P(W) + P(T|{\bar W}) P({\bar W}) $. So if you are interested in conditional events like $P(T|W)$, you could use this formula as follows. The additional info is that $ P(T|W) = 2 P(T|{\bar W}) $. So you obtain $0.3 = P(T|W) 0.4 + \frac12 P(T|W) 0.6$. This allows you to calculate $P(T|W) = 3/7$.

Now the situation $P*$ that it rains on any of the two days can be split into two disjoint events, so

$P* = P(W) + P(T|{\bar W}) P({\bar W}) = P(W) + \frac12 P(T|W) P({\bar W})= 0.4 + \frac12 \frac37 0.6 = \frac{37}{70} \simeq 0.528$

$p$ is the probability that it rains on Thursday if it has rained on Wednesday. $q$ is the probability that it rains on Thursday if it has not rained on Wednesday.

Given that the probability it rains on Thursday is $30\%$, we have $$ \frac25p+\frac35q=\frac3{10}\tag1 $$ Given that it is twice as likely to rain on Thursday if it has rained on Wednesday, we have $$ \frac{p}{1-p}=2\frac{q}{1-q}\tag2 $$ Solving $(1)$ and $(2)$ simultaneously, we get $$ p=\frac{17-\sqrt{193}}{8},q=\frac{-11+\sqrt{193}}{12}\tag3 $$ The probability that it rains on at least one day is the complement of the probability that it doesn't rain on either day. That is, $$ 1-\frac35(1-q)=\frac{-3+\sqrt{193}}{20}\approx54.4622\%\tag4 $$