2. If $x-y=1, x^{3}-y^{3}=2$, then $x^{4}+y^{4}=$
$\qquad$ $x^{5}-y^{5}=$ $\qquad$ .
2. $\frac{23}{9}, \frac{29}{9}$.

Since $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)=2, x-y=1$, we have
$$
x^{2}+x y+y^{2}=2 \text{. }
$$

Also, $x^{2}-2 x y+y^{2}=1$, combining this with the above equation, we get
$$
x y=\frac{1}{3}, x^{2}+y^{2}=\frac{5}{3} \text{. }
$$

Therefore, $x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=\frac{23}{9}$.
$$
\begin{array}{l}
\text{Also, } x^{5}-y^{5}=x^{5}-x^{4} y+x^{4} y-x y^{4}+x y^{4}-y^{5} \\
=x^{4}(x-y)+x y\left(x^{3}-y^{3}\right)+y^{4}(x-y),
\end{array}
$$

Substituting $x-y=1, x y=\frac{1}{3}, x^{3}-y^{3}=2$, we get
$$
x^{5}-y^{5}=x^{4}+y^{4}+\frac{2}{3}=\frac{29}{9} \text{. }
$$