(b)

We have,

$\log _{1 / 3}(x+y)+\log _3(x-y)=2$

$2 y^2=512^{x+1}$

$\Rightarrow \log _{3^{-1}}(x+y)+\log _3(x-y)=2$

$\Rightarrow-\log _3(x+y)+\log _3(x-y)=2$

$\Rightarrow \quad \frac{x-y}{x+y}=3^2=9$

$\Rightarrow \quad x-y=9 x+9 y$

$\Rightarrow \quad-8 x=10 y \Rightarrow-4 x=5 y$

and $\quad 2^{y^2}=2^{9(x+1)}$

$\Rightarrow \quad y^2=9(x+1)$

$\begin{array}{cc}\Rightarrow & 16 x^2=225 x+225 \\ \Rightarrow & 16 x^2-225 x-225=0 \\ \Rightarrow & (16 x+15)(x-15)=0 \\ \Rightarrow & x=15,-\frac{15}{16}\end{array}$

$x=-\frac{15}{16}, y=\frac{3}{4}$ (not possible)

$\therefore$ Only one solution $(15,-12)$.