What is details meaning of output and input Impedance?
To give a mathematical definition, if we were to excite the input of a two-port with a sinusoidal current source 

$$i_{in}(t) = I_0 + I \sin(\omega t)$$

with \$I\$ small enough to not cause any nonlinear behavior, we'd find

$$v_{in}(t) = V_0 + V_i \sin(\omega t) + V_q \cos(\omega t)$$

Then we could define

$$R_{in} = \frac{V_i}{I}$$

and

$$X_{in} = \frac{V_q}{I}.$$

Then the input impedance would be \$Z_{in}\$ if we defined

$$Z_{in} = R_{in} + j X_{in}.$$

In words, this means that the real part of \$Z_{in}\$ tells us how the in-phase component of the input voltage depends on the input current. And the imaginary part of \$Z_{in}\$ tells us how the quadrature (90 degree phase shifted) part of the input voltage depends on the input current.

We would also find that

$$R_{in} = \frac{{\rm{d}}V_0}{{\rm{d}}I_0}.$$

The output impedance is defined the same way, but as a relationship between the output current and output voltage.

All the other things we know about input and output impedance and how they relate to circuit behavior can be connected back to these definitions.


  "Maximum power transfer occurs when load impedance is equal to source impedance."


This isn't quite correct. 

First, when talking about possibly complex-valued impedances, maximum power transfer occurs when the source impedance is the complex conjugate of the load impedance.

Second, this maximum power transfer condition is for choosing the load impedance when the source impedance is fixed. If source impedance can be controlled, then either very high impedance (with a fixed-current source) or very low impedance (for a fixed-voltage source) should be chosen to deliver maximum power to a fixed impedance load.