Solving constants in the recursive term with master theorem
If you are just interested in an upper bound you can notice that $T(n) \le S(n)$ where $S(n) = 3 S(n/3) + \frac{n}{2}$ and has solution $S(n) = O(n \log n)$.
Alternatively there is always induction. You can show that, for $n \ge 2$,  $T(n) \le c n \log n$.
For $2 \le n &lt; 7$, $T(n)$ is a constant and $n \log n \ge 1$. Therefore the claim is true for a sufficiently large (constant) value $c^*$ of $c$.
For $n \ge 7$ you have:
$$
T(n) = 3T\left(\frac{n}{3} - 2\right) + \frac{n}{2} \le 3 c \frac{n}{3} \log \frac{n}{3} + \frac{n}{2}
= cn \log n - cn \log 3 + \frac{n}{2},
$$
which is at most $cn \log n$ when $c n \log 3 \ge \frac{n}{2}$ or, equivalently, $c \ge \frac{1}{2 \log 3}$.
Simply pick $c = \max\{c^*, \frac{1}{2 \log 3} \}$.