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Because $a^2-b^2 = (a-b)(a+b)$ , then we have the system $a+b = 2, \; a-b=1$ . This yields $a = \frac{3}{2}, \; b= \frac{1}{2}$ . Thus, $a^4 - b^4 = \frac{3^4-1}{2^4} = \frac{80}{16} = \boxed{5.}$

Please note $a^0-b^0 = 0$ is a tautology.

$\begin{aligned} a-b&=1 \\ a^2-b^2&=2 \\ \implies a+b&=2 \\ \end{aligned}$

$\text{Solving } a-b=1 \text{ and } a+b=2,$

$\begin{aligned} a&=\dfrac{3}{2} \\ b&=\dfrac{1}{2} \\ \implies a^4-b^4&=\boxed{5} \\ \end{aligned}$