Find $y^{(y+x)}$ ?

$175 = 5^{2}7^{1}, 245 = 5^{1}7^{2} \Rightarrow x =1, y = 2.$ Thus, $y^{x+y} = 2^{1+2} = 2^{3} = \boxed{8}.$

We can divide the two equations

$\frac{7^{y}•5^{x}}{7^{x}•5^{y}} = \frac{245}{175}$

$\frac{7^{y}•7^{-x}}{5^{y}•5^{-x}} = 1.4$

$\frac{7^{y-x}}{5^{y-x}} = 1.4$

$(\frac{7}{5})^{y-x} = 1.4$

$(1.4)^{y-x} = 1.4$

Implying $y-x= 1$ or $y= x+1$

From the first equation

$7^{x}•5^{y=x+1} = 175$

$(7×5)^{x} = 175/5$

$35^{x} = 35$

Meaning $x=1$ hence $y= 1+1= 2$

$y^{y+x}= 2^{2+1}= \boxed{8}$