Two identical straight wires are stretched so as to produce 8 beats...

The frequency of a stretched wire is given by $f = \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}}$.

Since $T_1 > T_2$, the initial frequencies satisfy $f_1 > f_2$.

The initial beat frequency is $f_1 - f_2 = 8$.

For the beat frequency to remain unchanged after changing the tensions, the new frequencies must satisfy $|f_1' - f_2'| = 8$.

If $T_1$ is changed while $T_2$ is kept constant, we need $|f_1' - f_2| = 8$. Since $f_1 - f_2 = 8$, the only other possibility is $f_2 - f_1' = 8 \Rightarrow f_1' = f_2 - 8 = f_1 - 16$. This requires $f_1$ to decrease, meaning $T_1$ must be decreased.

If $T_2$ is changed while $T_1$ is kept constant, we need $|f_1 - f_2'| = 8$. Since $f_1 - f_2 = 8$, the only other possibility is $f_2' - f_1 = 8 \Rightarrow f_2' = f_1 + 8 = f_2 + 16$. This requires $f_2$ to increase, meaning $T_2$ must be increased.

Thus, the beat frequency can remain unchanged if $T_1$ is decreased or $T_2$ is increased.

Answer: $T_1$ is decreased or $T_2$ is increased