(i) $\frac{1}{4}+\frac{1}{16} x-\frac{71}{64} x^2- \frac{71}{256} x^3$

(ii) $b=-0.280(3$ s.f.) $\quad a=-1.12(3$. s.f.)

(i) $(4-x)^{-1} \cos 3 x$
$=4^{-1}\left(1-\frac{1}{4} x\right)^{-1} \cos 3 x$

$\approx \frac{1}{4}\left(1-\left(-\frac{1}{4} x\right)+\frac{-1(-2)}{2 !}\left(-\frac{1}{4} x\right)^2+\frac{-1(-2)(-3)}{3 !}\left(-\frac{1}{4} x\right)^3\right)\left(1-\frac{(3 x)^2}{2 !}\right)$

$=\frac{1}{4}\left(1+\frac{1}{4} x+\frac{1}{16} x^2+\frac{1}{64} x^3\right)\left(1-\frac{9}{2} x^2\right)$
$\approx \frac{1}{4}\left(1-\frac{9}{2} x^2+\frac{1}{4} x-\frac{9}{8} x^3+\frac{1}{16} x^2+\frac{1}{64} x^3\right)$

$=\frac{1}{4}\left(1+\frac{1}{4} x-\frac{71}{16} x^2-\frac{71}{64} x^3\right)$
$=\frac{1}{4}+\frac{1}{16} x-\frac{71}{64} x^2-\frac{71}{256} x^3$

(ii) $(a+b x)^5=a^5+5 a^4 b x+10 a^3 b^2 x^2+10 a^2 b^3 x^3+\ldots b^5 x^5$

$10 a^3 b^2=-\frac{71}{64}$ — (1)

$10 a^2 b^3=-\frac{71}{256} \cdots(2)$ — (2)

(1) $\div(2), \frac{a}{b}=\frac{256}{64}=4$
$\therefore a=4 b$

Substituting into (1)
$10(4 b)^3 b^2 =-\frac{71}{64} $
$640 b^5 =-\frac{71}{64} $
$b^5 =-\frac{71}{40960} $
$b=-0.280(3$ s.f.) $\quad a=-1.12(3$. s.f.)