= Solution
For the first <linear map>, take $(\alpha,\beta)=(0,e_n)$. Its input <norm> is one, while $T_1(0,e_n)=(e_n,0)$ has <norm> $n^2$. Thus \b[$T_1$ is unbounded].
For the second map, since $j^2\geq1$,
$$
\|T_2(\alpha,\beta)\|_X=\|\alpha+\beta\|_1\leq\|\alpha\|_1+\|\beta\|_1\leq\|\alpha\|_{1,2}+\|\beta\|_1=\|(\alpha,\beta)\|_X.
$$
Equality holds at $(0,e_1)$, so
$$
\boxed{T_2\text{ is bounded and }\|T_2\|=1.}
$$
Both formulas map $c_{00}\times c_{00}$ into itself because adding two <finitely supported sequences> preserves finite support.
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