Both questions are essentially applications of the Central Limit Theorem, which says (informally) that "the value of a sum over many samples from a common population will tend to a normal distribution as the number of samples becomes large".

The two questions differ in the type of data that they treat. The "xbar" question concerns temperature, which is a continuous measurement (e.g. a decimal number). The "phat" question implicitly concerns a binary measurement (true/false, e.g. each student either invests or does not).

Commonly a measurement of a random variable will be denoted by $x$. For a random sample $x_1,\ldots,x_N$ the sample mean will then be denoted by $\bar{x}=\frac{1}{N}\sum_ix_i$. This applies directly to the "xbar" question. Here each $x_i$ is a temperature measurement, and the question asks about the sampling distribution of $\bar{x}$. (This arises when $\bar{x}$ is computed many times over different samples, each of size $N$).

For the "phat" question, the notation and logic is consistent with this, but the connection is a little more involved. In this case each $x_i$ will correspond to an individual student, who either invests ($x=1$) or does not ($x=0$). The probability that a student will invest would commonly be denoted by $p$ ($=30\%$ in this case). These conventions of $\Pr[\text{true}]=p$ and $\{\text{true,false}\}=\{1,0\}$ are standard for the case of a binary random variable.

Now imagine we do not know the value of $p$, but wish to estimate it from a random sample of students $x_1,\ldots,x_N$. For a single student the expected value of $x_i$ is $p$, denoted $\mathbb{E}[x]=p$ (see also here). Similarly, by the properties of expectation, for the sample we have $\mathbb{E}[\bar{x}]=p$. So here the sample mean $\bar{x}$ provides an estimate of the population parameter $p$. In statistics it is standard practice to denote an estimate of a population parameter by using a "hat", so here we it makes sense to denote the sample mean as $\hat{p}$.

(For the "xbar" problem the comparable notation would be $\bar{x}=\hat{\mu}$, as there $x$ is normal rather than Bernoulli.)