= Solution
Put $C=M_0^{1-\theta}M_1^\theta$. Since $p<\infty$, choose <simple functions> $f_n$ of finite-measure support with $f_n\to f$ in $L^p$. These belong to both endpoint domain spaces. The given simple-function estimate applies also to their differences, so
$$
\|Tf_n-Tf_m\|_q\leq C\|f_n-f_m\|_p.
$$
By <completeness of Lp spaces>, including $L^\infty$, there is $u\in L^q$ with $Tf_n\to u$ in $L^q$. Taking limits of the simple-function bound yields $\|u\|_q\leq C\|f\|_p$.
It remains to identify $u$ with the already defined $Tf$, rather than just define an unrelated extension. Write $X=L^{p_0}+L^{p_1}$ and $Y=L^{q_0}+L^{q_1}$ with their sum <norms>. The endpoint estimates imply
$$
\|Tv\|_Y\leq\max(M_0,M_1)\|v\|_X.
$$
Indeed, split $v=v_0+v_1$, bound the two images in their endpoint <norms>, and take the infimum over all splits. Part (i) gives $\|f_n-f\|_X\leq2\|f_n-f\|_p$, hence $Tf_n\to Tf$ in $Y$.
The inclusion $L^q\subseteq Y$ is also continuous. For distinct finite endpoints this is the same threshold argument as part (i), after ordering the endpoints. If the larger endpoint is infinity, splitting at $a=\|v\|_q$ bounds both the lower endpoint <norm> of the large-value part and the <supremum norm> of the small-value part by $\|v\|_q$. If the endpoints coincide, the inclusion is immediate. Thus $Tf_n\to u$ also in $Y$. Uniqueness of limits in the sum <norm> gives $u=Tf$ <almost everywhere>. Consequently the <Riesz-Thorin theorem> estimate extends as
$$
\boxed{\|Tf\|_q\leq M_0^{1-\theta}M_1^\theta\|f\|_p
\quad\text{for every }f\in L^p.}
$$
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