Evaluating the limit of $\lim_{n \to \infty} \frac{4n^2 +9}{2n^2+2n+3}$ with definition I have troubles with "estimating".
We want to evaluate the limit with the definition.  For all $\epsilon >0$ we have to find an $N$, such that $|a_n-a|<\epsilon$ holds for $n>N$.
The limit in question is:
$$\lim_{n \to \infty} \frac{4n^2 +9}{2n^2+2n+3}$$
We know the limit is $2$.

My approach:
Let $\epsilon > 0$.
$$\begin{align*} 
|a_n-2| 
&= \left|\frac{4n^2+9}{2n^2+2n+3} -2\right| \\
&= \left| \frac{4n^2+9 - 2(2n^2+2n+3)}{2n^2+2n+3} \right| \\
&= \left| \frac{4n^2+9 - 4n^2 - 4n -6}{2n^2+2n+3} \right| \\
&= \left| \frac{-4n+3}{2n^2+2n+3} \right| \\
&\leq \left| \frac{-4n}{2n^2+2n+3}\right|  \\
&\leq \frac{4n}{2n^2}  \\
&= \frac{2}{n} < \epsilon
\end{align*}$$
If we choose $N = \frac{2}{\epsilon}$ it holds that $\forall \epsilon > 0: |a_n -a| < \epsilon$, if $n > N$.

My professors approach:
$$\begin{align*}
\left| \frac{-4n+3}{2n^2+2n+3} \right| 
&\leq \left|\frac{3-4n}{2n^2} \right| \\
&\leq \left| \frac{3}{2n^2} \right| + \left|\frac{4n}{2n^2}\right| \\
&= \frac{3}{2} \cdot \frac{1}{n^2} + 2 \cdot \frac{1}{n}
\end{align*}$$
We note that:
$$\begin{align*}
&\frac{3}{2} \cdot \frac{1}{n^2} < \frac{\epsilon}{2} \Leftrightarrow \frac{3}{n^2} < \epsilon \iff n > \sqrt{\frac{1}{\epsilon}} \\
&2 \cdot \frac{1}{n}  < \frac{\epsilon}{2} \Leftrightarrow 4 \cdot \frac{1}{n} < 2 \iff \frac{4}{\epsilon} < n
\end{align*}$$
Then:
$$\begin{align*}
&\text{Let } N > \max\bigg\{\sqrt{\frac{1}{\epsilon}}, \frac{4}{\epsilon} \bigg\}  \\
&\implies\forall n > N \text{ it holds that } |a_n-a| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon
\end{align*}$$

I have the following questions:
$1.$ What is wrong with my estimation?
$2.$ How does my prof get to $\frac{3}{2} \cdot \frac{1}{n^2} < \frac{\epsilon}{2}$
$3.$ What exactly does $\max\bigg\{\sqrt{\frac{1}{\epsilon}}, \frac{4}{\epsilon} \bigg\}$ tell us?

Your working has no errors as far as I can tell. Sometimes proofs like these have multiple ways to approach it; this is one such instance.

To get to the line about $\newcommand{\ve}{\varepsilon}
\ve/2$, consider momentarily: we have two terms at this point in the working,
$$\frac 3 2 \cdot \frac{1}{n^2} \text{ and } 2 \cdot \frac 1 n$$
If $n$ is large enough, such that
$$\frac 3 2 \cdot \frac{1}{n^2} < \frac \ve 2 \text{ and } 2 \cdot \frac 1 n < \frac \ve 2$$
then clearly
$$\frac 3 2 \cdot \frac{1}{n^2} + 2 \cdot \frac 1 n < \frac \ve 2 + \frac \ve 2 = \ve$$
which is enough for what we want.

"Okay, but why bother?" It's just somewhat of a common trick in analysis to make each term in a $\ve-N$ or $\ve-\delta$ proof work individually, and then bring it all together in the end to make them all work together. It's a good trick to keep in mind, especially at points where the triangle inequality applies, which is something your professor used in getting to this point.

The goal, then, is to see what values of $n$ ensure this in each case. Well, solving for $n$, these respectively give
$$\begin{align*}
\frac 3 2 \cdot \frac{1}{n^2} < \frac \ve 2
&\implies \frac{1}{n^2} < \frac \ve 2 \cdot \frac 2 3 = \frac \ve 3 \\
&\implies n^2 >\frac 3 \ve \\
&\implies n > \sqrt{ \frac 3 \ve } > \sqrt{ \frac 1 \ve } = \frac{1}{\sqrt \ve} \tag{$\ast$} \\
2 \cdot \frac 1 n < \frac \ve 2
&\implies \frac 1 n < \frac \ve 4 \\
&\implies n > \frac 4 \ve \\
\end{align*}$$

Note: The choice to go to $1/\sqrt \ve$ as in $(\ast)$ is purely aesthetic, and serves no real purpose. (I certainly can't think of a meaningful reason why to do it, and I don't see why your instructor didn't do likewise for the latter equality and have $n > 1/\ve$ there.)

Anyhow. We know that if we choose $n>1/\sqrt \ve$, the first inequality is given, and if $n>4/\ve$, the second one is too. But we want both to be satisfied together. So we want $n$ greater than both of them.

Note: This ties precisely into "what does this tell us?" It tells us we broke down a necessary inequality up term by term, optimized each individually, and that each member of the $\max \{\cdots\}$ function makes (at least) one of those terms work.

The $\max$ function is ideal for this purpose, and ensures both are satisfied. It's as simple as realizing that
$$\begin{align*}
&\max \{x,y\} \ge x \\
&\max \{x,y\} \ge y
\end{align*}$$
All in all, this - optimizing individual terms, and using $\max \{\cdots\}$ to let it all work - might seem clunky at first, but it's a bread-and-butter trick for analysis. It's usually a lot easier to handle individual terms, rather than complicated fractions.
That doesn't invalidate your method, of course, but it would be a very good idea to be aware of this trick for future use.
