how can I generate Output 2 to give a fixed delay pulse on a rising and falling edge of Input?
Interpretation of OP's question.

                     ________________________
Input    ___________|                        |___________________
                     ________________________
Output 1 ___________|                        |___________________
                     __                       __
Output 2 ___________|  |_____________________|  |________________


Figure 1. System timing diagram.

Specification:


Input / output: 2.5 - 3.3 V DC at &lt; 40 mA.








simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. A possible solution.

How it works:


NAND1 to 4 are CMOS Schmitt trigger NAND gates. When both inputs are high the output goes low. Three of them are wired with the inputs tied together so these work as inverters.
On power up NAND3's inputs are pulled high by R1 and R2. When both inputs are high the output, OUT2, goes low.
NAND4 is a double-inverted version of the input signal. This satisfies the requirement for OUT1.
NAND1 and NAND2 provide an inverted and a double-inverted version of the input signal. We will use one to detect rising edges on the input and the other for falling.
When NAND1 output goes low (input going high) the negative going output causes the right-hand side of C1 to go to 0 V by capacitive coupling. This turns off one of the inputs of NAND3 so it's output goes high.
After a delay set by C1 x R1 (about 0.5 s) R1 charges C1 back up to 3.3 V and OUT2 goes low again. D1 protects NAND4 from overvoltage.
The circuit around NAND2 works the same way.


Cautions: 


Your requirement for inputs as low as 2.5 V may be on the edge of the CD4093 working correctly. Also the minimum operating voltage of the TI CD4093B is 3 V which is very close to your 3.3 V supply. When run on a 4 V supply anything above 2.2 V is read as a high and below 0.9 V is a low. This means it should work for your application. Try it and see.
The chip can only source or sink about 0.5 mA at that operating voltage. You will need to buffer the output. You haven't given the information on the source signal or the load so it isn't possible to offer advice on this.