(i) $-4<x<\frac{1}{2}$

(ii) $-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$

(i)

$ \frac{x+3}{x+4}  \leq \frac{5}{1-2 x}, x \neq-4, \frac{1}{2}$

      $\frac{x+3}{x+4}-\frac{5}{1-2 x} \leq 0$

      $\frac{(x+3)(1-2 x)-5(x+4)}{(x+4)(1-2 x)}  \leq 0$

      $\frac{-2 x^2-10 x-17}{(x+4)(1-2 x)}  \leq 0 $

     $\frac{-2\left(x^2+5 x+\frac{17}{2}\right)}{-(x+4)(2 x-1)} \leq 0$

     $\frac{2\left(\left(x+\frac{5}{2}\right)^2+\frac{9}{4}\right)}{(x+4)(2 x-1)} \leq 0$

Since $\left(x+\frac{5}{2}\right)^2-\frac{9}{A} \rtimes 0$ for all $x \in \mathbb{R}$,
$\frac{2}{(x+4)(2 x-1)} \leq 0$

$-4<x<\frac{1}{2}$

(ii)$\frac{x^2+3}{x^2+4} \leq \frac{5}{1-2 x^2}$
Replace $x$ with $x^2$ in inequality in (i), $-4<x^2<\frac{1}{2}$
Since $x^2 \geq 0,0 \leq x^2<\frac{1}{2} \Rightarrow x^2-\frac{1}{2}<0$
$$\begin{array}{l}
\left(x-\frac{1}{\sqrt{2}}\right)\left(x+\frac{1}{\sqrt{2}}\right)<0 \\
\therefore-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}
\end{array}
$$