Assume x={a}.
(a) $\bigcup${x} = x for all x.
1. Doesn't this mean {{a}}$\bigcup${{a}}$\bigcup${{a}} . . . $\bigcup${{a}} which would equal {a}?
(b) {$\bigcup$x} = x iff x is a singleton.
2. Wouldn't that expand to {{a}$\bigcup${a}$\bigcup${a} . . . $\bigcup${a}} which would end up being {{a}} which is not equal to {a}=x?
3. How and why is (b) true?
Thank you for your time and help!
On
Make sure to use the correct definitions. In this case it's
Let $X$ be a set (containing sets), then $$\bigcup X := \{ y : \exists Y\in X: y\in Y\}$$
From this we can prove the given statements. Let $x$ be a set. Then
$$\bigcup \{x\} = \{ y :\exists Y\in\{x\}:y\in Y\}$$
There is only one $Y$ in $\{x\}$, namely $Y=x$. We see $x$ as a set, therefore
$$\bigcup \{x\} = \{ y : y\in x\} = x$$
For (b):
Assume $\{\bigcup x\}=x$. Now $\bigcup x$ is just a set, say $a=\bigcup x$, so $\{a\} = x$, but then $x$ is already a singleton. Now assume $x$ is a singleton, i.e. $\{a\}=x$. But we know from (a) that, if we interpret $a$ as a set (instead of $x$), we can write $a=\bigcup\{a\}=\bigcup x$, so $\{\bigcup x\}=x$.
Your assumption that $x = \{a \}$ is irrelevant and actually harmful here. For any set $x$ we have that $\bigcup \{ x \} = x$.
Why? $\bigcup \{x \} = \{ z \mid \exists y \in \{x \} \colon z \in y \} = \{ z \mid z \in x \} = x$ (the last equality follows by extensionality).
Now, if $x$ is a singleton, i.e. $x = \{a \}$, then, by what we've shown above, $\{ \bigcup x \} = \{ \bigcup \{a \} \}= \{a \} = x$.
I'll leave it to you to show the converse. (If $x$ contains two distinct elements, then $\{ \bigcup x \} \neq x$.