The answer to question 1 is `yes': Let $A$ and $B$ be any C-algebras. Let $N$ be a simple C-algebra of such high cardinality that it does not embed into either $A$ or $B$. Then take $C:=N\oplus\mathbb{C}$.

We have $$ A\otimes C \cong (A\otimes N) \oplus A, \quad B\otimes C \cong (B\otimes N) \oplus B. $$ Let $\varphi\colon (A\otimes N) \oplus A\to (B\otimes N) \oplus B$ be an isomorphism. The choice of $N$ ensures that no quotient of $A\otimes N$ embeds into $B$. Thus, $\varphi$ maps the summand $A\otimes N$ to $B\otimes N$. Applying the same argument to the inverse of $\varphi$, we see that $\varphi$ is given by an isomorphism $A\otimes N\cong B\otimes N$ and an isomorphism $A\cong B$.

This also answers question 2 if one puts some restrictions on the cardinality (or better: density character) of the C-algebras. For instance, for separable C-algebras, one could use the test-algebra $C:=\mathcal{R}\oplus\mathbb{C}$, where $\mathcal{R}$ is the hyperfinite II-1 factor.

The answer to question 1 is `yes': Let $A$ and $B$ be any $C^*$-algebras. Let $N$ be a simple $C^*$-algebra of such high cardinality that it does not embed into either $A$ or $B$. Then take $C:=N\oplus\mathbb{C}$.

We have $$ A\otimes C \cong (A\otimes N) \oplus A, \quad B\otimes C \cong (B\otimes N) \oplus B. $$ Let $\varphi\colon (A\otimes N) \oplus A\to (B\otimes N) \oplus B$ be an isomorphism. The choice of $N$ ensures that no quotient of $A\otimes N$ embeds into $B$. Thus, $\varphi$ maps the summand $A\otimes N$ to $B\otimes N$. Applying the same argument to the inverse of $\varphi$, we see that $\varphi$ is given by an isomorphism $A\otimes N\cong B\otimes N$ and an isomorphism $A\cong B$.

This also answers question 2 if one puts some restrictions on the cardinality (or better: density character) of the $C^*$-algebras. For instance, for separable $C^*$-algebras, one could use the test-algebra $C:=\mathcal{R}\oplus\mathbb{C}$, where $\mathcal{R}$ is the hyperfinite II-1 factor.