I found the following identities about Pauli matrices from the lecture notes of Supersymmetry.

$$((\sigma^{\mu})^{\alpha\dot{\alpha}})^{\ast}=(\bar{\sigma}^{\mu})^{\dot{\alpha}\alpha} \tag{1}$$

$$((\sigma_{\mu\nu})^{\alpha}_{\,\,\,\beta})^{\ast}=-(\bar{\sigma}_{\mu\nu})_{\dot{\beta}}^{\,\,\,\dot{\alpha}} \tag{2}$$

$$((\sigma_{\mu\nu})^{\alpha\beta})^{\ast}=(\bar{\sigma}_{\mu\nu})^{\dot{\alpha}\dot{\beta}} \tag{3}$$

Please give me some hints on how to derive the above identities.

For the first identity, if we take $$\mu=0$$, then $$\sigma^{0}\equiv 1$$, which is real. Then, the identity says

$$(1^{\alpha\dot{\alpha}})^{\ast}=-1^{\dot{\alpha}\alpha}.$$

Why does the complex conjugation acting on this identity matrix $$1_{2\times 2}$$ exchange the two spinorial indices? Where does the minus sign come from?

Remember that $$\bar{\sigma}^{\mu \dot{a}b}=(1,-\vec{\sigma})^{\dot{a}b}$$
$$(\sigma^{\mu b \dot{a}})^*=\bar{\sigma}^{\mu \dot{a}b}$$
Taking $$\mu=0$$, would simply yield $$1=1$$ from the definition.
In any convention only one of $$\sigma^\mu$$ and $$\bar\sigma_\mu$$ can have both its spinor indices up. But in the original post above, in eq. (1) (which is as given on page 81 of the lecture notes) both $$\sigma^\mu$$ and $$\bar\sigma_\mu$$ have been shown with their spinor indices up. There are some typos. Eq. (1) should read as
$$(\sigma^{\mu\alpha\dot\beta})^*=\sigma^{\mu\beta\dot\alpha}\,.$$