You're right about inclusion and exclusion.
For the number of elements in exactly one of the sets, you first count all $n(A) + n(B) + n(C)$ elements. You disregard the $n(A \cap B)$ elements twice, as you need to disregard them from your count for both $A$ and $B$. The logic is similar for $A \cap C$ and $B \cap C$. Finally, you've disregarded the $n(A \cap B \cap C)$ elements in all three sets twice, so you have to correct the count by adding this quantity back in.
I bet you can figure out the second one once you've carefully understood the first.
It is not true for most sets $A$, because the elements of $A\times A$ are ordered pairs, and an ordered pair of elements from $A$ is not a usually set of elements of $A$ (which it would have to be in order to be a member of $\mathcal P(A)$.
In axiomatic set theory, where sets are the only things that exist, an ordered pair must be represented by a particular set -- the most common convention is to consider $(a,b)$ to "really" be an abbreviation for $\{\{a\},\{a,b\}\}$.
When this convention is used, there are a few particular cases where $A\times A$ is indeed a subset of $\mathcal P(A)$.
One example of this is $A=\varnothing$, in which case $A\times A=\varnothing$ which is a subset of anything. A different example is $A=H_\omega$, the set of hereditarily finite sets (which means basically the sets that can be written in finite space with only the symbols {, }, and ,, such as "$\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}$").
But in both of those cases $A\times A\subseteq \mathcal P(A)$ is more of an accident than something that tells us something interesting and useful about the sets in question.
Best Answer
Suppose $x_1$ and $x_2$ are distinct elements of $X$, $y_1$ and $y_2$ are distinct elements of $Y$.
If $X$ and $Y$ are disjoint, you can define $f\colon X\cup Y\to X\times Y$ by $$ \begin{cases} f(x)=(x,y_1) & \text{if $x\in X$, $x\ne x_1$}\\ f(x_1)=(x_1,y_2) \\ f(y)=(x_2,y) & \text{if $y\in Y$, $y\ne y_1$}\\ f(y_1)=(x_1,y_1) \end{cases} $$ This function is injective. Two distinct elements in $X$ have distinct images and two distinct elements in $Y$ have distinct images. So we can check whether it's possible that an element in $X$ and one in $Y$ have the same image: suppose $f(x)=f(y)$; if $x=x_1$, we have $f(x)=(x_1,y_2)$, but $f(y)=(x_2,y)$ or $f(y)=(x_1,y_1)$ and equality is impossible. So $x\ne x_1$ and $f(x)=(x,y_1)$. But, again $f(y)=(x_2,y)$ or $f(y)=(x_1,y_1)$; the second case is impossible, so we should have $y\ne y_1)$ and so $f(y)=(x_2,y)\ne(x,y_1)=f(x)$.
Suppose now $X$ and $Y$ are not disjoint and write $X\cup Y=X\cup(Y\setminus X)$.
If $Y\setminus X$ has at least two elements, we get an injective map $X\cup Y\to X\times(Y\setminus X)\subseteq X\times Y$ by the argument above. Otherwise $Y\setminus X$ is empty or it has just one element.
If $Y\setminus X=\emptyset$, we have $Y\subseteq X$ and we can define $f\colon X\to X\times Y$ by $f(x)=(x,y_1)$ which is injective.
If $Y\setminus X=\{y_0\}$ we can define $f\colon X\cup\{y_0\}\to X\times Y$ by $f(x)=(x,y_0)$ for $x\in X$ and $f(y_0)=(x_1,y_1)$ if $y_0\ne y_1$ or $f(y_0)=(x_1,y_2)$ if $y_0=y_1$.