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Fix some calculation error.
Zhanxiong
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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 $\Phi$.

Zhanxiong
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