For part (a), you can show linearity like this:
$$
\langle\langle x + y, z \rangle\rangle = \operatorname{Re}(\langle x + y, z \rangle) = \operatorname{Re}(\langle x, z \rangle + \langle y, z \rangle) = \operatorname{Re}(\langle x, z \rangle) + \operatorname{Re}(\langle y, z \rangle) = \langle\langle x, z \rangle\rangle + \langle\langle y, z \rangle\rangle
$$
$$
\langle\langle cx, y \rangle\rangle = \operatorname{Re}(\langle cx, y \rangle) = \operatorname{Re}(c\langle x, y \rangle) = c\operatorname{Re}(\langle x, y \rangle) = c\langle\langle x, y \rangle\rangle
$$
Note that $c \in \mathbb{R}$, so the third equality in the second line is valid.
For part (b), all you need to say is that $\langle x, x \rangle \in \mathbb{R}$ even in complex inner products since it follows from the conjugate symmetric axiom. So taking the real part doesn't change the value and so $\langle\langle x, x \rangle\rangle = \langle x, x \rangle$. Now take square roots to show norms are equal.
For (c), you can clearly see that if the complex inner product is zero then taking the real part is still zero which proves the first part. But it is possible for the complex inner product to be non-zero but its real part zero (i.e. the value only has non-zero imaginary part). So the converse is false.
Disclaimer: Throughout the discussion, I'll always assume that normed spaces are real or complex (and preferably real).
The existence of some $y$ in $C$ such that $\lVert x-y\rVert=\min_{z\in C} \lVert x-z\rVert$ is a consequence of the fact that closed balls are compact. There must be some closed ball $E(x,R)$ such that $E(x,R)\cap C\ne \emptyset$.
Since $C$ is closed and $E(x,R)$ is compact, $E(x,R)\cap C$ is compact; therefore, there is some $y\in E(x,R)\cap C$ that minimizes the continuous function $\lVert x-\bullet\rVert$ on $E(x,R)\cap C$. Now, it is easy to observe that $\min\left\{\lVert x-z\rVert\,:\, z\in C\right\}=\min\left\{\lVert x-z\rVert\,:\, z\in C\cap E(x,R)\right\}$. For, if $z\notin E(x,R)$, then $\lVert x-z\rVert$ is automatically larger than $R$ and thus of the distance from $x$ of any element of $E(x,R)\cap C$.
So $y$ is indeed a minimizer such as the ones you want. Without additional hypothesis uniqueness won't be guaranteed, basically for the same reason it isn't guaranteed in general. If $V=(\Bbb R^2,\lVert\bullet\rVert_\infty)$ and $C=\{1\}\times [-1,1]$ and $x=(0,0)$, then $d(x,y)=1$ for all $x\in C$.
Following this line of thought, you may want to prove that a normed space is strictly convex if and only if it has the property that for all closed convex sets $C$ and $x\notin C$ there is at most one $y$ such that $\lVert x-y\rVert=\min\{\lVert x-z\rVert\,:\, z\in C\}$.
Best Answer
It is the norm of the orthogonal projection of $v$ on $F$.