Answer
See below
Work Step by Step
a) Obtain: $Ker (T)=\{p \in P_2(R): T(p)=0\}\\
=\{ax^2+bx+c:(a+b)+x(b-c)=0\}$
We have the system:
$a+b=0\\
b-c=0$
thus, $a=-b\\
c=b$
Thus, $Ker (T)=\{-bx^2+bx+b:b\in R\}\\
=\{b(-x^2+x+1):b\in R\}\\
=span\{-x^2+x+1\}$
$\dim [Ker(T)]=1$
b) Apply Rank Nullity Theorem:
$\dim [Ker(T)]+\dim [Rng(T)]=\dim P_2(R)\\
1+\dim [Rng(T)]=3\\
\dim [Rng(T)=2$
Since $Rng(T) \subset P_1(R)\\
\dim [Rng(T)]=\dim P_2(R)\\
\rightarrow Rng(T)=P_1(R)$
