(a) $\frac{x^2}{4}+y^2=1$
Differentiating w.r.t x,
$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{2x}{4(2y)}=-\frac{x}{4y}$
The gradient of the curve is
$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{4(\frac{\sqrt{3}}{2})}$
= $-\frac{1}{2\sqrt{3}}$
Equation of the tangent at $(1,\frac{\sqrt{3}}{2})$ is
$\frac{y-\frac{\sqrt{3}}{2}}{x-1}=-\frac{1}{2\sqrt{3}}$
$1-x=2\sqrt{3}(y-\frac{\sqrt{3}}{2})$
$1-x=2\sqrt{3}y-3⟹2\sqrt{3}y+x-4=0$.
(b) $\frac{3x+2}{x^2+x-2}$
$x^2+x-2=x^2-x+2x-2$
$x(x-1)+2(x-1)⟹x^2+x-2\equiv (x-1)(x+2)$
$\frac{3x+2}{x^2+x-2}=\frac{A}{x-1}+\frac{B}{x+2}$
= $\frac{A(x+2)+B(x-1)}{(x-1)(x+2)}$
Comparing with the equation given, we have
$3x+2=A(x+2)+B(x-1)$
= $Ax+2A+Bx-B$
$⟹A+B=3...(1)$
$2A-B=2...(2)$
(1) + (2) : $3A=5⟹A=\frac{5}{3}$
$\frac{5}{3}+B=3⟹B=3-\frac{5}{3}=\frac{4}{3}$
$\therefore \frac{3x+2}{x^2+x-2}=\frac{5}{3(x-1)}+\frac{4}{3(x+2)}$.