Why it is necessary that phase of incident, reflected and refracted wave must equal at the interface of two medium? 1.Why it is necessary that phase of incident, reflected and  refracted wave must equal at the interface of two medium to satisfy the boundary conditions at the interface?
2. According to boundary conditions fields left to the interface must join fields right to interface why we just add amplitudes of fields E(I) + E (R) = E(T) here  E(I), E (R) and E(T) are amplitudes of incident,reflected and transmitted wave 
 A: It's not necessary that the phase should be equal, as in the case of $v_2<v_1$, the reflected and incident wave is out of phase.
What is important that the field must satisfy the electrodynamic boundary conditions :
$$(1) \ \ \epsilon_1E_1^\perp=\epsilon_2E^\perp_2, \ \ \ \ \ (2) \ \ \ \ E^\parallel_1=E_2^\parallel$$
$$(3) \ \ \ B_1^\perp=B_2^\perp, \ \ \ \ \  \ (4) \ \ \ \  \ \frac{1}{\mu_1}B^\parallel_1=\frac{1}{\mu_2}B^\parallel_2$$
These equations relate the electric and magnetic fields just to the left and just to the right of the interface between two linear media.
The rest follows from here.

Additional Answer On Edited Question:
It's not necessary to use Amplitude, You can, if want to, use the fields. Suppose like as in Griffith, the fields are :
$$\mathbf{E}_I(z,t)=E_{0I}e^{i(k_1z-\omega t)}\hat{x}$$
$$\mathbf{E}_R(z,t)=E_{0R}e^{i(-k_1z-\omega t)}\hat{x}$$
$$\mathbf{E}_T(z,t)=E_{0T}e^{i(k_1z-\omega t)}\hat{x}$$
The interface is at $z=0$ so
$$\mathbf{E}_I(0,t)+\mathbf{E}_R(0,t)=\mathbf{E}_T(0,t)$$
$$\Rightarrow E_{0I}e^{i(k_1\cdot 0-\omega t)}+E_{0R}e^{i(-k_1\cdot 0-\omega t)}=E_{0T}e^{i(k_1\cdot 0-\omega t)}$$
$$E_{0I}+E_{0R}=E_{0T}$$
That's the same thing as to add amplitudes.
