Chances of raining over the weekend, chance of rain on saturday is 25% and chance of rain on sunday is 25% Right now I'm torn between using logic and using Bayer's thingy.
My logic tells me:
Change of rain on saturday : 25%
Chance of rain on sunday : 25%
Chance of rain on sunday and saturday 0.25²
Which gets me at around 56% chance of rain. Using Bayes would get me at around 60 i think
Is my thinking correct? 
 A: Your thinking is not correct ... assuming you are asked to find the chance of it raining anytime over the weekend.
Assuming the event of it raining on Saturday is independent of it raining on Sunday (which is a pretty dubious assumption, but hey!), the chance of it raining over the weekend is the chance of it raining on Saturday or on Sunday ... which is 1 minus the chance of it not raining on both days, which is:
$$1-(0.75)^2=\frac{7}{16}$$
We can also use:
$$P(SaturdayRain \cup SundayRain) =$$
$$ P(SaturdayRain) + P(SundayRain) \color{red}{-} P(SaturdayRain \cap SundayRain) =$$
$$ 0.25 +0.25 - (0.25)^2 \approx 0.44$$
I think you tried to use this formula but added the probability of it raining on both days rather than subtracting it, thus ending up with about $56$%. But as pointed out in the comments, it of course makes no sense if the chance of either one happening are greater than the chances of either one happening individually added together.
A: The compliment rule is the most simple approach here.
$$\begin{align*}
P(\text{rains this weekend})
&= 1-P(\text{does not rain this weekend}) \\\\
&= 1-\frac{3}{4}^2 \\\\
&= .4375
\end{align*}$$
A: Let $A$ be the event it rains on Saturday and $U$ be the event it rains on Sunday. If it rains over the weekend, it rains on Saturday or Sunday or both; let this be the event $E=A\cup U$.
The easiest way to calculate $\operatorname{P} (E)$ is $$\operatorname{P} (E) = 1-\operatorname{P} \!\left(E’\right) = 1-\operatorname{P}\!\left(A’\right)\!\,\operatorname{P} \!\left(U’\right)\!$$
Can you take it from there?
A: \begin{eqnarray*}
P(\text{Rain})=1-P( \text{no rain}) =1-P((\text{ no rain on sat) and ( no rain on sun)} ) \\= 1-P( \text{no rain sat}) P ( \text{no rain sun}) =1- \frac{3}{4}\times \frac{3}{4} =\frac{7}{16}.
\end{eqnarray*}
So thats $43.75 \%$ chance of rain. You double counted it raining on both days. 
