$T(\phi)=T(\psi)$ if $\phi$ and $\psi$ agree on the support of $T$? I just started learning about distributions. I learned what the support of a distribution $T$ is. It is the complement of the biggest set $E$, such that for every test function $\phi$ on the set $E$ we have $T(\phi)=0$ (is this right). Now I have a question. If $\phi, \psi$ are two test functions that are different but agree on the support of the distribution $T$, then $T(\phi)=T(\psi)$ right? Although I observed it is true for distributions like $T(\phi)=\int f\phi dx$, I don't know how to prove that in general case.

The support is rather the complementary of the biggest open set $E$ such that ... (the rest is correct). This small change implies that the support is always a closed set.
The answer to your question is yes. I will rather work with $\phi_1$ and $\phi_2$. Since these two functions agree on the support $\Omega$ of $T$, we can write $\phi_1 = \psi + \tilde \phi_1$ and $\phi_2 = \psi + \tilde \phi_2$ where $\tilde \phi_i \equiv 0$ in $\Omega$ (using a smooth partition of unity over a set eventually slightly bigger than $\Omega$). Thus for both $i$ :
$$T(\phi_i)=T(\psi + \tilde \phi_i)=T(\psi)+T(\phi_i) = T(\psi).$$
