Why is the Voltage only 1.4 for reset input when in high impedance?
You are missing a 10 K pull-up resistor in series with a 1N4148 diode at the reset pin of the MPU itself. You show CRESET as being there for programming the IC, but no 3.3 volt pull up which MUST be there. Your reset transistor needs a current on its collector at all times or the IC may stay in reset state.
The diode is to prevent 12 volts from going back into the Vcc 3.3 volt line during programing.
Pin 1 of your DB25 connector should have a 10 K resistor to ground so the reset transistor does not turn ON due to leakage currents.
To reset the IC you need to ground its RESET pin or connect pin1 of the DB25 connector to Vcc. At least that should be your goal.
I am not sure where or why you get 1.4 volts on the reset pin. First of all your are not defining which reset pin. The IC or the DB25 pin 1. Plus you are missing important pull-up and pull-down resistors so by my standards the 1.4 volts is just leakage current.
I am sure you will see the 1.4 volt reading change and be more sensible once you install the resistors and diode I mentioned.
You missed a couple of important steps that should get it working correctly, at least the RESET function anyways.
In all honesty your diagram is a bit confusing. I am only looking at the left schematic with the DB25 connector, which you have pin1 as RESET if it is high. It cannot be open collector unless it is a PNP transistor because T1 must have its base pulled low or it will reset the MPU.
You are sooo close to having it right. Yes the diode is forward biased from a 3.3 or 5 volt line, through the 10 K resistor. This insures that the MPU RESET pin is held high, yet can still act as a programming pin. The 3 pin connector should NOT be used unless you are programming the MPU.