= Solution
For scalars $\alpha,\beta$ and vectors $v,w,u$, use the adjoint identity and conjugate linearity in the second argument:
$$
\begin{aligned}
\langle u,T^*(\alpha v+\beta w)\rangle
&=\langle Tu,\alpha v+\beta w\rangle\\
&=\overline\alpha\langle Tu,v\rangle+\overline\beta\langle Tu,w\rangle\\
&=\overline\alpha\langle u,T^*v\rangle+\overline\beta\langle u,T^*w\rangle\\
&=\langle u,\alpha T^*v+\beta T^*w\rangle.
\end{aligned}
$$
The difference of the two second-argument vectors is orthogonal to every $u$. Choosing $u$ to be that difference shows its squared <norm> is zero. Therefore
$$
\boxed{T^*(\alpha v+\beta w)=\alpha T^*v+\beta T^*w.}
$$
The two conjugations are why the adjoint is a linear map, despite the conjugate-linear dependence of the representing functional on its second argument.
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