= Solution
{c}
For any column vector $v$ and <positive semidefinite> input $\varrho$, the <Kraus representation> gives
$$
v^\dagger\mathcal E(\varrho)v=\sum_k(A_k^\dagger v)^\dagger\varrho(A_k^\dagger v)\geq0.
$$
Thus the output is <positive semidefinite>. The <cyclic property of the trace> and completeness of the <Kraus operators> give
$$
\boxed{\operatorname{tr}\mathcal E(\varrho)=\operatorname{tr}\!\left(\varrho\sum_kA_k^\dagger A_k\right)=\operatorname{tr}\varrho.}
$$
More strongly, tensoring with an arbitrary <ancilla> replaces $A_k$ by $A_k\otimes I$, so the same argument proves <complete positivity> rather than only positivity on single-system <density matrices>.
If $\mathcal F$ has <Kraus operators> $B_j$, the <composition of quantum channels> has
$$
(\mathcal F\circ\mathcal E)(\varrho)=\sum_{j,k}B_jA_k\varrho A_k^\dagger B_j^\dagger,
\qquad
\sum_{j,k}(B_jA_k)^\dagger B_jA_k
=\sum_kA_k^\dagger\!\left(\sum_jB_j^\dagger B_j\right)A_k=I.
$$
This is again a trace-preserving <Kraus representation>. The composition of <completely positive maps> is <completely positive> even without the trace-preserving assumption; in the general case, the product <Kraus operators> still establish <complete positivity>.
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