Solution: 1. Let us assume that such a function exists. That is, there exists a function $f:\mathbb{C} \setminus \{0\} \to \mathbb{C} \setminus \{0\}$ such that for every $z,w\in \mathbb{C} \setminus \{0\}$ \begin{equation}\label{eqn:25Jan-01} f(zw)=f(z)f(w),~\text{ and }~f \left( z^{2} \right) =z. \end{equation}

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Now we set $z=w=1$, then using \eqref{eqn:25Jan-01}, we have \begin{align*} &f(1) = f(1)^2 \text{ and } f(1) = 1 \\ \implies & f(1) = f(1)^2 = 1. \end{align*} Similarly, we set $z=w=-11$, then using \eqref{eqn:25Jan-01}, we have \begin{align*} &f(1) = f(-1)^2 \text{ and } f(1) = -1 \\ \implies & f(1) = f(-1)^2 = -1, \end{align*} which is a contradiction. Hence, such a function does not exist.

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2. Again, we assume that such a function exists. That is, there is a function $f:\mathbb{C} \setminus \{0\}\to \mathbb{C} \setminus \{0\}$ such that \[ f(z)^2=z,~\forall~z\in \mathbb{C} \setminus \{0\}. \] Let $z_0\in \mathbb{C} \setminus \{0\}$ and consider the following function. \[ g(z) = \frac{f\left( z_0z \right) }{f(z)f\left( z_0 \right) }. \]

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Since $f$ is continuous and never takes the value $0$, so $g$ is continuous on $\mathbb{C} \setminus \{0\}$. Observe that \begin{align*} g(z)^2 & = \left( \frac{f\left( z_0z \right) }{f(z)f\left( z_0 \right) } \right) ^2 \\ & = \frac{f\left( z_0z \right) ^2}{f(z)^2f\left( z_0 \right)^2 }\\ & = \frac{z_0z}{zz_0} \\ & = 1. \end{align*} Thus, \[ g(z) = \pm 1. \]

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As $g$ is a continuous function on $\mathbb{C} \setminus \{0\}$ and $\mathbb{C} -\{0\}$ is connected set, therefore the image must be connected (continuous image of connected space is connected ). Therefore, \[ g(z) \equiv 1, \text{ or } g(z) \equiv -1. \] If $g(z)\equiv 1$, then we have \begin{align*} f(z)f(z_0) = f\left( zz_0 \right). \end{align*} In particular, \[ f\left( z_0 \right) ^2 = f\left( z_0^2 \right), \forall~z_0\in \mathbb{C} \setminus \{0\}. \] Since the above is true for all $z_0$, so \[ f(z)^2 = f\left( z^2 \right)\implies f\left( z^2 \right) = z^2. \] Using part (a), this function will not exist and hence we got a contradiction. Therefore, $g(z)\equiv 1$ is not possible.

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Now take $g(z) \equiv -1$. Then we have \[ f(z)f\left( z_0 \right) = -f\left( zz_0 \right). \] If we replace $f$ by $-f$, then we will get a contradiction same as above. So, $g(z)\equiv -1$ is also not possible. Thus, we proved that such a function does not exist.