Answer
See below
Work Step by Step
Assume $v \in V$
We know that $\{v_1,v_2\}$ is a basic for the vector space $V$ there exist scalars $\alpha, \beta$ such as:
$$v=\alpha v_1+\beta v_2$$
Obtain: $T_1T_2(v)=(T_1T_2)(\alpha v_1+\beta v_2)\\
=T_1(T_2(\alpha v_1+\beta v_2)) \\
=T_1(\alpha T_1(v_1)+\beta T_2(v_2)\\
=T_1(\alpha .\frac{1}{2}(v_1+v_2)+\beta .\frac{1}{2}(v_1-v_2))\\
=T_1((\frac{1}{2}\alpha +\frac{1}{2}\beta)v_1+(\frac{1}{2} \alpha -\frac{1}{2}\beta)v_2)\\
=\frac{1}{2}(\alpha +\beta)T_1(v_1)+\frac{1}{2}(\alpha -\beta)T_1(v_2)\\
=\frac{1}{2}(\alpha +\beta)(v_1+v_2)+\frac{1}{2}(\alpha-\beta)(v_1-v_2)\\
=[\frac{1}{2}(\alpha +\beta)+\frac{1}{2}(\alpha-\beta)]v_1+[\frac{1}{2}(\alpha +\beta)-\frac{1}{2}(\alpha -\beta)]v_2\\
=(\frac{1}{2}\alpha +\frac{1}{2}\beta+\frac{1}{2}\alpha -\frac{1}{2}\beta)v_1+(\frac{1}{2}\alpha +\frac{1}{2}\beta-\frac{1}{2}\alpha+\frac{1}{2}\beta)v_2\\
=\alpha v_1+\beta v_2\\
=v$
then $T_2T_1(v)=(T_2T_1)(\alpha v_1+\beta v_2)\\
=T_2(T_1(\alpha v_1+\beta v_2)) \\
=T_2(\alpha T_1(v_1)+\beta T_1(v_2)\\
=T_2(\alpha (v_1+v_2)+\beta (v_1-v_2))\\
=T_2((\alpha +\beta)v_1+(\alpha -\beta)v_2)\\
=(\alpha +\beta)T_2(v_1)+(\alpha -\beta)T_2(v_2)\\
=\frac{1}{2}(\alpha +\beta)(v_1+v_2)+\frac{1}{2}(\alpha-\beta)(v_1-v_2)\\
=[\frac{1}{2}(\alpha +\beta)+\frac{1}{2}(\alpha-\beta)]v_1+[\frac{1}{2}(\alpha +\beta)-\frac{1}{2}(\alpha -\beta)]v_2\\
=(\frac{1}{2}\alpha +\frac{1}{2}\beta+\frac{1}{2}\alpha -\frac{1}{2}\beta)v_1+(\frac{1}{2}\alpha +\frac{1}{2}\beta-\frac{1}{2}\alpha+\frac{1}{2}\beta)v_2\\
=\alpha v_1+\beta v_2\\
=v$
$T_2T_1(v)= \\
T_1T_2(v)\\
\rightarrow [(T_1^{-1}T_1)(v)]=T_2T_1(v)=v\\
\rightarrow T_2=T_1^{-1}\\
$
