Asymptotic bound $T(n)=T(n/3+\lg n)+1$ How would I go about finding the upper and lower bounds of $T(n)=T(n/3+\lg(n))+1$?

Not sure how tight a bound you need, but here is an idea.
Compare your recurrence to the one that satisfies $X_n = X_{n/3} + 1$ (without the log) - what is the relationship between $T_n$ and $X_n$? Note that when you solve, $X_n = \Theta(\log n)$.
Next item is in the other direction. Compare $T_n$ to $Y_n = Y_{n-1}+1$, and note that $Y_n = \Theta(n)$.
If you fill in the blanks, you get an inequality bounding on both sides.
