Evaluate:

(i) $3^{-4}=\frac{1}{3^{4}}=\frac{1}{81}$

(ii) $(-4)^{3}=(-1)^{3} \times(4)^{3}=-1 \times 64=-64$

(iii) $\left(\frac{3}{4}\right)^{-2}=\left(\frac{4}{3}\right)^{2}=\frac{4^{2}}{3^{2}}=\frac{16}{9}$

(iv) $\left(\frac{-2}{3}\right)^{-5}=\left(\frac{3}{-2}\right)^{5}=\frac{3^{5}}{-2^{5}}=\frac{243}{-32}=\frac{243 \times-1}{-32 \times-1}=\frac{-243}{32}$

(v) Using the property $\left(\frac{a}{b}\right)^{0}=1$, we have:

$\left(\frac{5}{7}\right)^{0}=1$