Answer
See below
Work Step by Step
Assume $x=(x_1,x_2)\\
y=(y_1,y_2)\\
c\in R$
$T(x+y)=T((x_1,x_2)+(y_1,y_2))\\
=T(x_1+y_1,x_2+y_2)\\
=\frac{x_1+y_1+x_2+y_2}{5}\\
=\frac{x_1+x_2}{5}+\frac{y_1+y_2}{5}\\
=T(x_1+x_2)+T(y_1+y_2)\\
=T(x)+T(y)$
$T(cf)=T(c(x_1,x_2))\\
=T(cx_1,cx_2)\\
=\frac{cx_1+cx_2}{5}\\
=c(\frac{x_1+x_2}{5})\\
=cT(x_1,x_2)\\
=cT(x)$
Hence, $T$ is a linear transformation
Obtain: $Ker(T)=\{x \in R^2:T (x)=0\}\\
=\{(x,y)\in R^2: T(x,y)=(0,0)\}\\
=\{(x,y)\in R^2: \frac{x+y}{5}\\
=\{(x,y):x+y=0\}\\
=\{(x,y):y=-x\}\\
=\{(x,-x)\}\\
=\{x(1,-1)\}\\
=span \{(1,-1)\}\\
\rightarrow \dim Ker(T)=1$
then $\dim [Ker(T)]=1 \ne \{0\}$
$T$ is not one-to-one
Apply Rank Nullity Theorem:
$\dim [Ker(T)]+\dim [Rng(T)]=\dim R^3\\
1+\dim [Rng(T)]=2\\
\dim [Rng(T)]=1$
Since $Rng(T) \subseteq R$ and $\dim [Rng(T)]=1=\dim R$,
hence, $T$ is onto.
Consequently, basic for $Rng(T)$ is $\{1\}$
