Consider the following recurrence relations: For all $n > 1$, $$T_1(n) = 4T_1\left(\frac{n}{2}\right) + T_2(n)$$ $$T_2(n) = 5T_2\left(\frac{n}{4}\right) + \Theta(\log_2 n)$$ Assume that for all $n \le 1$, $T_1(n) = 1$ and $T_2(n) = 1$. Which one of the following options is correct?

First, solve the recurrence for $T_2(n)$ using the Master Theorem:

Here, $a = 5$, $b = 4$, and $f(n) = \Theta(\log_2 n)$.

Compare $f(n)$ with $n^{\log_b a} = n^{\log_4 5} \approx n^{1.161}$: Since $f(n) = O\left(n^{\log_4 5 - \epsilon}\right)$ for $\epsilon > 0$, by Case 1 of the Master Theorem:

Now substitute $T_2(n)$ into the recurrence for $T_1(n)$:

Here, $a = 4$, $b = 2$, and $f(n) = \Theta\left(n^{\log_4 5}\right)$.

Calculate $n^{\log_b a}$:

Since $\log_4 5 \approx 1.161 < 2$, we have $f(n) = O\left(n^{2 - \epsilon}\right)$ for some $\epsilon > 0$.

By Case 1 of the Master Theorem:

Thus, the correct option is (A).