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No - take $(\mathcal{P}(\mathbb{N}), \tau)$ where $\tau = \{\emptyset\}\cup\{A\subseteq\mathbb{N}: \mathbb{N}\setminus A \text{ is finite}\}$. Clearly, every two non-empty open sets have non-empty intersection, so the space is hyperconnected, but is not path-connected, see this post.

No - take $(\mathcal{P}(\mathbb{N}), \tau)$ where $\tau = \{\emptyset\}\cup\{A\subseteq\mathbb{N}: \mathbb{N}\setminus A \text{ is finite}\}$. Clearly, every two non-empty open sets have non-empty intersection, so the space is hyperconnected, but is not path-connected, see this post.

No - take $(\mathcal{P}(\mathbb{N}), \tau)$ where $\tau = \{\emptyset\}\cup\{A\subseteq\mathbb{N}: \mathbb{N}\setminus A \text{ is finite}\}$. Clearly, every two non-empty open sets have non-empty intersection, so it the space is hyperconnected, but is not path-connected, see this post.

No - take $(\mathcal{P}(\mathbb{N}), \tau)$ where $\tau = \{\emptyset\}\cup\{A\subseteq\mathbb{N}: \mathbb{N}\setminus A \text{ is finite}\}$. Clearly, every two non-empty open sets have non-empty intersection, so the space is hyperconnected, but is not path-connected, see this post.

No - take $(\mathcal{P}(\mathbb{N}), \tau)$ where $\tau = \{A\subseteq\mathbb{N}: \mathbb{N}\setminus A \text{ is finite}\}$. Clearly, every two non-empty open sets have non-empty intersection, so it the space is hyperconnected, but is not path-connected, see this post.

No - take $(\mathcal{P}(\mathbb{N}), \tau)$ where $\tau = \{\emptyset\}\cup\{A\subseteq\mathbb{N}: \mathbb{N}\setminus A \text{ is finite}\}$. Clearly, every two non-empty open sets have non-empty intersection, so it the space is hyperconnected, but is not path-connected, see this post.