(A) Find the 95th percentile of a normal distribution with mean 100 and standard deviation 20.
So Z = $\frac {X- \mu}{\sigma}$. And were given $\mu$ and $\sigma$, so I'm assuming X is 95? If I take this as X,I get Z = -0.25, and from looking at a table the 95th quantile would be 0.4013. Am I thinking about this right?
(B) Find the 25th percentile of a normal distribution with mean 10.5 and variance 65.
Is the X here just 25/10.5?
$\bbox[pink,0.25ex]X$ is the random variable of interest and $\bbox[pink, 0.25ex]x$ is the unknown value you wish to find.
$X$ is presumed to be normally distributed with mean $\bbox[pink,0.25ex]\mu$ and variance $\bbox[pink,0.25ex]{\sigma^2}$. So we use the following : where $Z$ is a normally distributed random variable with mean $0$ and variance $1^2$, then:
$$\mathsf P(X\leq x) ~=~ \mathsf P(Z\leq \dfrac{x-\mu}{\sigma})$$
Now a value $\bbox[pink,0.25ex]\alpha$ for any $\bbox[pink,0.25ex]p$ (eg $p=0.95$, or $p=0.25$, et cetera) in $~\bbox[pink,0.25ex]{\mathsf P(Z\leq \alpha)= p}~$ can be found by looking up your Student's $Z$ tables. (So named as they were published by a mathematician using the alias "Student").
When you have found $\alpha$ from the tables, then clearly $\bbox[pink,0.25ex]{x = \alpha\sigma+\mu}$ gives you the value you really want.