Solving $T(n) = 3T(n-1) + 2$
Let us add $1$ to both sides of your recurrence:
$$
T(n) + 1 = 3T(n-1) + 3 = 3(T(n-1) + 1).
$$
This shows that
$$
T(n) + 1 = 3^{n-1}(T(1) + 1),
$$
and so
$$
T(n) = 2 \cdot 3^{n-1} - 1.
$$