Simplify the following complex number expressions: a) (2+3i)/(i(4-5i)) + 2 b) 1/(2+3i) + 1/(1-2i)

a)

$$\frac{2+3i}{i(4-5i)} + 2$$ $$= \frac{2+3i}{4i-5i^2} + 2$$ $$= \frac{2+3i}{5+4i} + 2$$

Multiply by the conjugate of the denominator:

$$= \frac{2+3i}{5+4i} \times \frac{5-4i}{5-4i} + 2$$ $$= \frac{(2)(5) + (2)(-4i) + (3i)(5) + (3i)(-4i)}{5^2+4^2} + 2$$ $$= \frac{10 - 8i + 15i - 12i^2}{25+16} + 2$$ $$= \frac{10 + 7i + 12}{41} + 2$$ $$= \frac{22+7i}{41} + 2$$ $$= \frac{22}{41} + \frac{7}{41}i + \frac{82}{41}$$ $$= \frac{22+82}{41} + \frac{7}{41}i$$ $$= \frac{104}{41} + \frac{7}{41}i$$

The final answer is $\boxed{\frac{104}{41} + \frac{7}{41}i}$.

b)

$$\frac{1}{2+3i} + \frac{1}{1-2i}$$

Rationalize each fraction:

$$= \frac{1}{2+3i} \times \frac{2-3i}{2-3i} + \frac{1}{1-2i} \times \frac{1+2i}{1+2i}$$ $$= \frac{2-3i}{2^2+3^2} + \frac{1+2i}{1^2+2^2}$$ $$= \frac{2-3i}{4+9} + \frac{1+2i}{1+4}$$ $$= \frac{2-3i}{13} + \frac{1+2i}{5}$$

Find a common denominator (65):

$$= \frac{5(2-3i)}{65} + \frac{13(1+2i)}{65}$$ $$= \frac{10-15i + 13+26i}{65}$$ $$= \frac{(10+13) + (-15+26)i}{65}$$ $$= \frac{23+11i}{65}$$ $$= \frac{23}{65} + \frac{11}{65}i$$

The final answer is $\boxed{\frac{23}{65} + \frac{11}{65}i}$.