jmc

Since ${\left(\frac{a}{b}\right)}^n={\left(\frac{b}{a}\right)}^{-n}$, then we know ${\left(\frac{3}{4}\right)}^{-3}={\left(\frac{4}{3}\right)}^3$.

Also, we know ${\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n}$, so ${\left(\frac{4}{3}\right)}^3=\frac{4^3}{3^3}$ and ${\left(\frac{1}{2}\right)}^8=\frac{1^8}{2^8}$.

So, combining terms, we have ${\left(\frac{1}{2}\right)}^8\cdot {\left(\frac{3}{4}\right)}^{-3}=\frac{4^3}{3^3}\cdot \frac{1^8}{2^8}=\frac{4^3\cdot 1^8}{3^3\cdot 2^8}$. We can simplify since $4^3=64=2^6$, and $\frac{2^6}{2^8}=\frac{1}{2^2}$ (since $\frac{a^k}{a^j}=a^{k-j}$).

Then, we solve for $\frac{4^3\cdot 1^8}{3^3\cdot 2^8}=\frac{1}{3^3}\cdot \frac{2^6}{2^8}=\frac{1}{3^3}\cdot \frac{1}{2^2}$. Since $3^3=3\cdot 3\cdot 3=27$, and $2^2=4$, we plug in these values, and find our answer to be $\frac{1}{27}\cdot \frac{1}{4}=\frac{1}{27\cdot 4}=\boxed{\frac{1}{108}}$.