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

As answered already, the lower bound is $\Theta(\lg n)$. 
If you want a tighter upper bound, you can observe that for any $n>32$, $\lg(n) < n/6$. So if you replace $T(n/3+\lg n) + 1$ with $T(n/2)+1$, this behaves like $\Theta(\lg(n))$. Then, as $n$ gets less than $32$, $n/3 + \lg n$ still less than $n$, so the remaining part is bounded by a constant.
The upper bound is then $\Theta(\lg n)$ too, and your bounds are tight.
