For the problem "finding $c$ such that $P(Z^2I(Z > 0) > c) = \alpha$", it is entirely unnecessary to introduce any conditional notation or argument (please see my comment above too). It is just a simple application of the law of total probability.

To begin with, $c$ must be non-negative, for the random variable $Z^2I(Z > 0)$ is non-negative hence if $c$ is negative, then $\alpha = 1$, which is impractical (assuming $\alpha$ is the size of some hypothesis testing problem). That said, by the law of total probability: \begin{align*} & P(Z^2I(Z > 0) > c) \\ =& P(Z^2I(Z > 0) > c, Z > 0) + P(Z^2I(Z > 0) > c, Z \leq 0) \\ =& P(Z^2 > c, Z > 0) + P(0 > c, Z \leq 0) \\ =& P(Z > \sqrt{c}) = 1 - \Phi(\sqrt{c}). \end{align*} The last equality is due to $Z \sim N(0, 1)$. Solving $1 - \Phi(\sqrt{c}) = \alpha$ then gives $$c = (\Phi^{-1}(1 - \alpha))^2,$$ where $\Phi^{-1}(p)$ denotes the $p$-quantile of $\Phi$.

For the problem "finding $c$ such that $P(Z^2I(Z > 0) > c) = \alpha$", it is entirely unnecessary to introduce any conditional notation or argument (please see my comment above too). It is just a simple application of the law of total probability.

To begin with, $c$ must be non-negative, for the random variable $Z^2I(Z > 0)$ is non-negative hence if $c$ is negative, then $\alpha = 1$, which is impractical (assuming $\alpha$ is the size of some hypothesis testing problem). That said, by the law of total probability: \begin{align*} & P(Z^2I(Z > 0) > c) \\ ={}& P(Z^2I(Z > 0) > c, Z > 0) + P(Z^2I(Z > 0) > c, Z \leq 0) \\ ={}& P(Z^2 > c, Z > 0) + P(0 > c, Z \leq 0) \\ ={}& P(Z > \sqrt{c}) = 1 - \Phi(\sqrt{c}). \end{align*} The last equality is due to $Z \sim N(0, 1)$. Solving $1 - \Phi(\sqrt{c}) = \alpha$ then gives $$c = (\Phi^{-1}(1 - \alpha))^2,$$ where $\Phi^{-1}(p)$ denotes the $p$-quantile of $\Phi$.

For the problem "finding $c$ such that $P(Z^2I(Z > 0) > c) = \alpha$", it is entirely unnecessary to introduce any conditional notation or argument (please see my comment above too). It is just a simple application of law of total probability.

To begin with, $c$ must be non-negative, for the random variable $Z^2I(Z > 0)$ is non-negative hence if $c$ is negative, then $\alpha = 1$, which is impractical (assuming $\alpha$ is the size of some hypothesis testing problem). That said, by the law of total probability: \begin{align*} & P(Z^2I(Z > 0) > c) \\ =& P(Z^2I(Z > 0) > c, Z > 0) + P(Z^2I(Z > 0) > c, Z \leq 0) \\ =& P(Z^2 > c) + P(0 > c, Z \leq 0) = P(Z^2 > c). \end{align*} Since $Z \sim N(0, 1)$, $Z^2 \sim \chi_1^2$, whence $c$ satisfying $P(Z^2 > c) = \alpha$ should be the upper-$\alpha$ quantile of the $\chi_1^2$ distribution, i.e., $c = \chi_1^2(1 - \alpha)$.

For the problem "finding $c$ such that $P(Z^2I(Z > 0) > c) = \alpha$", it is entirely unnecessary to introduce any conditional notation or argument (please see my comment above too). It is just a simple application of the law of total probability.

To begin with, $c$ must be non-negative, for the random variable $Z^2I(Z > 0)$ is non-negative hence if $c$ is negative, then $\alpha = 1$, which is impractical (assuming $\alpha$ is the size of some hypothesis testing problem). That said, by the law of total probability: \begin{align*} & P(Z^2I(Z > 0) > c) \\ =& P(Z^2I(Z > 0) > c, Z > 0) + P(Z^2I(Z > 0) > c, Z \leq 0) \\ =& P(Z^2 > c, Z > 0) + P(0 > c, Z \leq 0) \\ =& P(Z > \sqrt{c}) = 1 - \Phi(\sqrt{c}). \end{align*} The last equality is due to $Z \sim N(0, 1)$. Solving $1 - \Phi(\sqrt{c}) = \alpha$ then gives $$c = (\Phi^{-1}(1 - \alpha))^2,$$ where $\Phi^{-1}(p)$ denotes the $p$-quantile of $\Phi$.