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The general recipe to construct such varieties is the following.

Start with your favorite smooth variety $Y$ such that $\operatorname{Pic}^0(Y) \neq 0$ (for instance, this condition is automatically satisfied if $H^0(Y, \, \Omega_Y^1) \neq 0$), and choose a non-zero, $2$-torsion divisor $\mathcal{L}$ in $\operatorname{Pic}(Y)$. Correspondingly, there is an étale double cover $$\pi \colon X \to Y, \quad \pi_*\mathcal{O}_X= \mathcal{O}_Y \oplus \mathcal{L},$$ and the generator of the deck transformations of $\pi$ is a fixed-point free holomorphic involution on $X$.

The Kodaira dimension can only increase under this procedure, namely $\operatorname{kod}(X) \geq \operatorname{kod}(Y)$. In particular, if we start with $Y$ a variety of general type (for example, a product of curves of genus $\geq 2$), it follows that $X$ is of general type, too. Thus, the answer to your second question is no.

The general recipe for constructing such varieties is the following.

Start with your favorite smooth variety $Y$ such that $\operatorname{Pic}^0(Y) \neq 0$ (for instance, this condition is automatically satisfied if $H^0(Y, \, \Omega_Y^1) \neq 0$), and choose a non-zero, $2$-torsion divisor $\mathcal{L}$ in $\operatorname{Pic}^0(Y)$. Correspondingly, there is an étale double cover $$\pi \colon X \to Y, \quad \pi_*\mathcal{O}_X= \mathcal{O}_Y \oplus \mathcal{L},$$ and the generator of the deck transformations of $\pi$ is a fixed-point free holomorphic involution on $X$.

The Kodaira dimension can only increase under this procedure, namely $\operatorname{kod}(X) \geq \operatorname{kod}(Y)$. In particular, if we start with $Y$ a variety of general type (for example, a product of curves of genus $\geq 2$), it follows that $X$ is of general type, too. Thus, the answer to your second question is no.

A simple recipe to construct such varieties is the following.

Start with your favorite smooth variety $Y$ such that $\operatorname{Pic}^0(Y) \neq 0$ (for instance, this condition is automatically satisfied if $H^0(Y, \, \Omega_Y^1) \neq 0$), and choose a non-zero, $2$-torsion divisor $\mathcal{L}$ in $\operatorname{Pic}(Y)$. Correspondingly, there is an étale double cover $$\pi \colon X \to Y, \quad \pi_*\mathcal{O}_X= \mathcal{O}_Y \oplus \mathcal{L},$$ and the generator of the deck transformations of $\pi$ is a fixed-point free holomorphic involution on $X$.

The Kodaira dimension can only increase under this procedure, namely $\operatorname{kod}(X) \geq \operatorname{kod}(Y)$. In particular, if we start with $Y$ a variety of general type (for example, a product of curves of genus $\geq 2$), it follows that $X$ is of general type, too. Thus, the answer to your second question is no.

The general recipe to construct such varieties is the following.

Start with your favorite smooth variety $Y$ such that $\operatorname{Pic}^0(Y) \neq 0$ (for instance, this condition is automatically satisfied if $H^0(Y, \, \Omega_Y^1) \neq 0$), and choose a non-zero, $2$-torsion divisor $\mathcal{L}$ in $\operatorname{Pic}(Y)$. Correspondingly, there is an étale double cover $$\pi \colon X \to Y, \quad \pi_*\mathcal{O}_X= \mathcal{O}_Y \oplus \mathcal{L},$$ and the generator of the deck transformations of $\pi$ is a fixed-point free holomorphic involution on $X$.

The Kodaira dimension can only increase under this procedure, namely $\operatorname{kod}(X) \geq \operatorname{kod}(Y)$. In particular, if we start with $Y$ a variety of general type (for example, a product of curves of genus $\geq 2$), it follows that $X$ is of general type, too. Thus, the answer to your second question is no.

A simple recipe to construct such varieties is the following.

Start with your favorite smooth variety $Y$ such that $\operatorname{Pic}^0(Y) \neq 0$ (for instance, this condition is automatically satisfied if $H^0(Y, \, \Omega_Y^1) \neq 0$), and choose a non-zero, $2$-torsion divisor $\mathcal{L}$ in $\operatorname{Pic}(Y)$. Correspondingly, there is an étale double cover $\pi \colon X \to Y$, and the generator of the deck transformations of $\pi$ is a fixed-point free holomorphic involution on $X$.

The Kodaira dimension can only increase with this procedure, namely $\operatorname{kod}(X) \geq \operatorname{kod}(Y)$. In particular, if we start with $Y$ a variety of general type (for instance, a product of curves of genus $\geq 2$), it follows that $X$ is of general type, too. Thus, the answer to your second question is no.

A simple recipe to construct such varieties is the following.

Start with your favorite smooth variety $Y$ such that $\operatorname{Pic}^0(Y) \neq 0$ (for instance, this condition is automatically satisfied if $H^0(Y, \, \Omega_Y^1) \neq 0$), and choose a non-zero, $2$-torsion divisor $\mathcal{L}$ in $\operatorname{Pic}(Y)$. Correspondingly, there is an étale double cover $$\pi \colon X \to Y, \quad \pi_*\mathcal{O}_X= \mathcal{O}_Y \oplus \mathcal{L},$$ and the generator of the deck transformations of $\pi$ is a fixed-point free holomorphic involution on $X$.

The Kodaira dimension can only increase under this procedure, namely $\operatorname{kod}(X) \geq \operatorname{kod}(Y)$. In particular, if we start with $Y$ a variety of general type (for example, a product of curves of genus $\geq 2$), it follows that $X$ is of general type, too. Thus, the answer to your second question is no.