1. (i) Solution:
$$-4 x+8 y i+7=2 x-3 y i+7 i$$

Rearranging terms
$$-6 x+11 y i=-7+7 i$$
Thus
$$x=\frac{7}{6}, \quad y=\frac{7}{11}$$
(ii) Solution:
$$x+y i=\sqrt{a+b i}$$

Squaring both sides gives
$$x^{2}-y^{2}+2 x y i=a+b i,$$

Therefore
$$\left\{\begin{array}{l} x^{2}-y^{2}=a \\ 2 x y=b \end{array}\right.$$

Squaring and adding the two equations gives

That is $\square$
$$\begin{array}{l} \left(x^{2}-y^{2}\right)^{2}+4 x^{2} y^{2}=a^{2}+b^{2} \\ \left(x^{2}+y^{2}\right)^{2}=a^{2}+b^{2} \\ x^{2}+y^{2}=\sqrt{a^{2}+b^{2}} \end{array}$$

Solving (1) and (3) gives
$$x^{2}=\frac{\sqrt{a^{2}+b^{2}}+a}{2}, y^{2}=\frac{\sqrt{a^{2}+b^{2}}-a}{2},$$

Thus
$$x= \pm \sqrt{\frac{\sqrt{a^{2}+b^{2}}+a}{2}}, y= \pm \sqrt{\frac{\sqrt{a^{2}+b^{2}}-a}{2}}$$

From (2), when $b>0$, $x$ and $y$ have the same sign, and when $b<0$, $x$ and $y$ have opposite signs.