Why is $\dfrac{1}{2}n^2-3n = \Theta(n^2)$?
Since $n&gt;0$, $\dfrac 3n &gt;0$. Hence $\dfrac{1}{2}-\dfrac{3}{n} \leq  c_2$ is implied by $\dfrac 12\le c_2$.

That is, your (2) is not correct while everything else is fine.

Just in case you cannot see why your (2) is wrong, please take a look at the following deduction.
$$ \dfrac{1}{2}-\dfrac{3}{n} \leq  c_2\\
\dfrac{1}{2}-c_2\le\dfrac{3}{n} \\
\dfrac{1/2-c_2}{3} \leq  \dfrac 1n\\
\dfrac 1n\ge\dfrac{1/2-c_2}{3} $$
Compare the last inequality with your (2).