Solve the equation (jx)/(1+jy) = (3x+j4)/(x+3y) and express z = (a+jb)/(c+jd) in the form X+jY.

$$jx^2 + j3xy = 3x + j3xy + j4 + j^24y$$

Since $j^2 = -1$, substitute this into the equation:

$$jx^2 + j3xy = 3x + j3xy + j4 - 4y$$

Rearrange the right side to group real and imaginary terms:

$$j(x^2 + 3xy) = (3x - 4y) + j(3xy + 4)$$

Equating the real parts:

$$0 = 3x - 4y \quad (*1)$$

Equating the imaginary parts:

$$x^2 + 3xy = 3xy + 4 \quad (*2)$$

From $(*2)$:

$$x^2 = 4$$ $$x = \pm 2$$

From $(*1)$, we have $3x = 4y$, so $y = \frac{3x}{4}$.

Case 1: If $x=2$

$$y = \frac{3(2)}{4} = \frac{6}{4} = \frac{3}{2}$$

Case 2: If $x=-2$

$$y = \frac{3(-2)}{4} = \frac{-6}{4} = -\frac{3}{2}$$

The solutions are:

$$\boxed{x=2, y=\frac{3}{2} \quad \text{and} \quad x=-2, y=-\frac{3}{2}}$$

$$z = \frac{a+jb}{c+jd} \times \frac{c-jd}{c-jd}$$ $$z = \frac{(a+jb)(c-jd)}{c^2 - (jd)^2}$$ $$z = \frac{ac - ajd + jbc - j^2bd}{c^2 - j^2d^2}$$

Since $j^2 = -1$:

$$z = \frac{ac - ajd + jbc + bd}{c^2 + d^2}$$

Group the real and imaginary parts:

$$z = \frac{(ac+bd) + j(bc-ad)}{c^2+d^2}$$ $$z = \frac{ac+bd}{c^2+d^2} + j\frac{bc-ad}{c^2+d^2}$$

Here, the real part is $X = \frac{ac+bd}{c^2+d^2}$ and the imaginary part is $Y = \frac{bc-ad}{c^2+d^2}$.

$$Y = 0$$ $$\frac{bc-ad}{c^2+d^2} = 0$$

Since $c^2+d^2 \neq 0$, the numerator must be zero:

$$bc-ad = 0$$ $$bc = ad$$

Divide both sides by $bd$ (assuming $b \neq 0$ and $d \neq 0$):

$$\frac{bc}{bd} = \frac{ad}{bd}$$ $$\boxed{\frac{c}{d} = \frac{a}{b}}$$

$$X = 0$$ $$\frac{ac+bd}{c^2+d^2} = 0$$

Since $c^2+d^2 \neq 0$, the numerator must be zero:

$$ac+bd = 0$$ $$ac = -bd$$

Divide both sides by $bc$ (assuming $b \neq 0$ and $c \neq 0$):

$$\frac{ac}{bc} = \frac{-bd}{bc}$$ $$\boxed{\frac{a}{b} = -\frac{d}{c}}$$

$$(1+j)^2 = 1^2 + 2(1)(j) + j^2$$ $$= 1 + 2j - 1$$ $$= 2j$$

$$j(2+j) = 2j + j^2$$ $$= 2j - 1$$

$$\text{RHS} = (2j) + (2j - 1)$$ $$\text{RHS} = -1 + 4j$$

Equating the real parts:

$$a+b = -1 \quad (*1)$$

Equating the imaginary parts:

$$a-b = 4 \quad (*2)$$

Add $(*1)$ and $(*2)$:

$$(a+b) + (a-b) = -1 + 4$$ $$2a = 3$$ $$a = \frac{3}{2}$$

Substitute $a = \frac{3}{2}$ into $(*1)$:

$$\frac{3}{2} + b = -1$$ $$b = -1 - \frac{3}{2}$$ $$b = -\frac{2}{2} - \frac{3}{2}$$ $$b = -\frac{5}{2}$$

The values of $a$ and $b$ are:

$$\boxed{a=\frac{3}{2}, b=-\frac{5}{2}}$$