Can Master Theorem be applied on any of these? 1) $T(n) = 6T(n/2) + 2^{3 \log(n)}$
2) $T(n) = 8T(n/2) + \frac{n^3}{(\log(n))^4}$
3) $T(n) = 9T(n/3) + n(\log(n))^3$
Can the complexity for these be calculated with the Master Theorem?
I am not sure how to decide upon which case they fit in.

For your second problem, lets solve the recurrence
$$T(n) = 8 T(\lfloor n/2 \rfloor) + 
\frac{n^3}{(1+\lfloor\log_2 n\rfloor)^4}$$
where $T(0) = 0.$
Let $$n = \sum_{k=0}^{\lfloor\log_2 n\rfloor} d_k 2^k$$ be the binary representation of $n.$
Unrolling the recursion we find the exact formula
$$T(n) = \sum_{j=0}^{\lfloor\log_2 n\rfloor} 
\frac{8^j}{(1+\lfloor\log_2 n\rfloor - j)^4}
\left(\sum_{k=j}^{\lfloor\log_2 n\rfloor} d_k 2^{k-j}\right)^3.$$
Now to get an upper bound on this consider the case of $n$ being a string of one digits, which gives
$$T(n)\le \sum_{j=0}^{\lfloor\log_2 n\rfloor} 
\frac{8^j}{(1+\lfloor\log_2 n\rfloor - j)^4}
\left(\sum_{k=j}^{\lfloor\log_2 n\rfloor} 2^{k-j}\right)^3
\\ = \sum_{j=0}^{\lfloor\log_2 n\rfloor} 
\frac{1}{(1+\lfloor\log_2 n\rfloor - j)^4}
\left(\sum_{k=j}^{\lfloor\log_2 n\rfloor} 2^k\right)^3
=  \sum_{j=0}^{\lfloor\log_2 n\rfloor} 
\frac{1}{(1+\lfloor\log_2 n\rfloor - j)^4} (2^{\lfloor\log_2 n\rfloor+1} - 2^j)^3
\\ = \sum_{j=1}^{\lfloor\log_2 n\rfloor+1}
\frac{1}{j^4} (2^{\lfloor\log_2 n\rfloor+1} - 2^{\lfloor\log_2 n\rfloor+1-j})^3
= 2^{3(\lfloor\log_2 n\rfloor+1)}
\sum_{j=1}^{\lfloor\log_2 n\rfloor+1}
\frac{1}{j^4} (1 - 2^{-j})^3.$$
The sum term converges rapidly to a constant and hence the asymptotics of the upper bound are
$$\frac{1}{90}
\left(\pi^4 -270 \mathrm{Li}_4(1/2) + 270 \mathrm{Li}_4(1/4) - 90 \mathrm{Li}_4(1/8)\right)
\times 2^{3(\lfloor\log_2 n\rfloor+1)}.$$
For the a lower bound consider the case of a one digit followed by zeros, which gives
$$T(n)\ge \sum_{j=0}^{\lfloor\log_2 n\rfloor} 
\frac{8^j}{(1+\lfloor\log_2 n\rfloor - j)^4}
\left(2^{\lfloor\log_2 n\rfloor-j}\right)^3
= 2^{3\lfloor\log_2 n\rfloor}\sum_{j=0}^{\lfloor\log_2 n\rfloor} 
\frac{1}{(1+\lfloor\log_2 n\rfloor - j)^4}
\\ = 2^{3\lfloor\log_2 n\rfloor}\sum_{j=1}^{\lfloor\log_2 n\rfloor+1} \frac{1}{j^4} .$$
The sum term once again converges to a constant and the asymptotics of the lower bound are
$$\frac{\pi^4}{90} 2^{3\lfloor\log_2 n\rfloor}.$$
Joining the two bounds we see that
$$T(n) \in \Theta\left(2^{3\lfloor\log_2 n\rfloor}\right) = \Theta(n^3).$$
