Express (2+j3)/(j(4-j5)) + 2/j in the form a+jb.

For the first term's denominator:

$$j(4-j5) = j4 - j^2 5 = j4 - (-1)5 = 5+j4$$

For the second term:

$$\frac{2}{j} = \frac{2}{j} \times \frac{-j}{-j} = \frac{-j2}{-j^2} = \frac{-j2}{-(-1)} = \frac{-j2}{1} = -j2$$

$$\frac{2+j3}{5+j4} = \frac{2+j3}{5+j4} \times \frac{5-j4}{5-j4}$$

Numerator:

$$(2+j3)(5-j4) = 2(5) + 2(-j4) + j3(5) + j3(-j4)$$ $$= 10 - j8 + j15 - j^2 12$$ $$= 10 + j7 - (-1)12$$ $$= 10 + j7 + 12 = 22 + j7$$

Denominator:

$$(5+j4)(5-j4) = 5^2 - (j4)^2$$ $$= 25 - j^2 16$$ $$= 25 - (-1)16$$ $$= 25 + 16 = 41$$

So the first term simplifies to $\frac{22+j7}{41} = \frac{22}{41} + j\frac{7}{41}$.

$$\left(\frac{22}{41} + j\frac{7}{41}\right) + (-j2)$$ $$= \frac{22}{41} + j\frac{7}{41} - j\frac{82}{41}$$ $$= \frac{22}{41} + j\left(\frac{7-82}{41}\right)$$ $$= \frac{22}{41} - j\frac{75}{41}$$

The expression in the form $a+jb$ is $\boxed{\frac{\text{22}}{\text{41}} - j\frac{\text{75}}{\text{41}}}$.

: If $z = \frac{1}{2+j3} + \frac{1}{1-j2}$, express $z$ in the form $a+jb$.

$$\frac{1}{2+j3} = \frac{1}{2+j3} \times \frac{2-j3}{2-j3}$$ $$= \frac{2-j3}{2^2 - (j3)^2} = \frac{2-j3}{4 - j^2 9} = \frac{2-j3}{4 - (-1)9} = \frac{2-j3}{4+9} = \frac{2-j3}{13} = \frac{2}{13} - j\frac{3}{13}$$

$$\frac{1}{1-j2} = \frac{1}{1-j2} \times \frac{1+j2}{1+j2}$$ $$= \frac{1+j2}{1^2 - (j2)^2} = \frac{1+j2}{1 - j^2 4} = \frac{1+j2}{1 - (-1)4} = \frac{1+j2}{1+4} = \frac{1+j2}{5} = \frac{1}{5} + j\frac{2}{5}$$

$$z = \left(\frac{2}{13} - j\frac{3}{13}\right) + \left(\frac{1}{5} + j\frac{2}{5}\right)$$

Combine the real parts and the imaginary parts:

$$z = \left(\frac{2}{13} + \frac{1}{5}\right) + j\left(-\frac{3}{13} + \frac{2}{5}\right)$$

Find a common denominator for the real parts (13 and 5, LCM is 65):

$$\frac{2}{13} + \frac{1}{5} = \frac{2 \times 5}{13 \times 5} + \frac{1 \times 13}{5 \times 13} = \frac{10}{65} + \frac{13}{65} = \frac{23}{65}$$

Find a common denominator for the imaginary parts (13 and 5, LCM is 65):

$$-\frac{3}{13} + \frac{2}{5} = \frac{-3 \times 5}{13 \times 5} + \frac{2 \times 13}{5 \times 13} = \frac{-15}{65} + \frac{26}{65} = \frac{11}{65}$$

So, $z = \frac{23}{65} + j\frac{11}{65}$.
The expression for $z$ in the form $a+jb$ is $\boxed{\frac{\text{23}}{\text{65}} + j\frac{\text{11}}{\text{65}}}$.