7.290. $\left\{\begin{array}{l}y^{x}=1.5+y^{-x} \\ y^{2.5+x}=64(y>0)\end{array}\right.$
## Solution.

Multiply the first equation by $y^{x}$, we have $y^{2 x}-1.5 y^{x}-1=0$. Solving this equation as a quadratic in terms of $y^{x}$, we get $y^{x}=-\frac{1}{2}$ (no solutions), or $y^{x}=2$. From the second equation of the system $y^{2.5} \cdot y^{x}=64 \Rightarrow$ $y^{2.5} \cdot 2=64, y^{2.5}=32, y=4$. Therefore, $4^{x}=2, x=\frac{1}{2}$.

$$
\text { Answer: }\left(\frac{1}{2} ; 4\right)
$$