No, this does not exist, unless you allow the trivial solution $I = \{0\}$. Otherwise let $\tau$ be a tracial state on $I$ and let $(e_\lambda)$ be a quasi-central approximate unit for $I$. This means that $e_\lambda x - xe_\lambda \to 0$ for all $x \in A$. Then $\phi: x \mapsto \lim \tau(e_\lambda x)$ is a tracial state on $A$.

The verification is straightforward. If $x \geq 0$ then $e_\lambda x - x^{1/2}e_\lambda x^{1/2} \to 0$, and $x^{1/2}e_\lambda x^{1/2}$ is increasing so $\tau(x^{1/2}e_\lambda x^{1/2})$ converges. This shows that $\tau(e_\lambda x)$ converges. Positivity of $\phi$ also follows because $\tau(x^{1/2}e_\lambda x^{1/2}) \geq 0$, and $\phi$ is a trace because, for any $x,y \in A$ we have $\tau(e_\lambda xy) = \tau(ye_\lambda x)$ and $ye_\lambda x - e_\lambda yx \to 0$.

No, this does not exist, unless you allow the trivial solution $I = \{0\}$. Otherwise let $\tau$ be a tracial state on $I$ and let $(e_\lambda)$ be a quasi-central approximate unit for $I$. This means that $e_\lambda x - xe_\lambda \to 0$ for all $x \in A$. Then $(e_\lambda^{1/2})$ is also an approximate unit and $\phi: x \mapsto \lim \tau(e_\lambda x)$ is a tracial state on $A$.

The verification is straightforward. If $x \geq 0$ then $e_\lambda x - x^{1/2}e_\lambda x^{1/2} \to 0$, and $x^{1/2}e_\lambda x^{1/2}$ is increasing so $\tau(x^{1/2}e_\lambda x^{1/2})$ converges. This shows that $\tau(e_\lambda x)$ converges. Positivity of $\phi$ also follows because $\tau(x^{1/2}e_\lambda x^{1/2}) \geq 0$, and $\phi$ is a trace because, for any $x,y \in A$ we have $\tau(e_\lambda xy) \approx \tau(e_\lambda^{1/2} xe_\lambda^{1/2} y) = \tau(e_\lambda^{1/2} ye_\lambda^{1/2} x) \approx \tau(e_\lambda yx)$, where $\approx$ means the difference goes to zero in norm.

No, this does not exist, unless you allow the trivial solution $I = \{0\}$. Otherwise let $\tau$ be a tracial state on $I$ and fix $x \in I$ such that $a = \tau(x^*x) > 0$. Then $y \mapsto \frac{1}{a}\tau(x^*yx)$ is a tracial state on $A$.

No, this does not exist, unless you allow the trivial solution $I = \{0\}$. Otherwise let $\tau$ be a tracial state on $I$ and let $(e_\lambda)$ be a quasi-central approximate unit for $I$. This means that $e_\lambda x - xe_\lambda \to 0$ for all $x \in A$. Then $\phi: x \mapsto \lim \tau(e_\lambda x)$ is a tracial state on $A$.

The verification is straightforward. If $x \geq 0$ then $e_\lambda x - x^{1/2}e_\lambda x^{1/2} \to 0$, and $x^{1/2}e_\lambda x^{1/2}$ is increasing so $\tau(x^{1/2}e_\lambda x^{1/2})$ converges. This shows that $\tau(e_\lambda x)$ converges. Positivity of $\phi$ also follows because $\tau(x^{1/2}e_\lambda x^{1/2}) \geq 0$, and $\phi$ is a trace because, for any $x,y \in A$ we have $\tau(e_\lambda xy) = \tau(ye_\lambda x)$ and $ye_\lambda x - e_\lambda yx \to 0$.