The statement of Hahn-Banach is
Given a normed vector space $X$ and a subspace $Y \subseteq X$, then every linear continuous $y^* \in Y^*$ can be extended to the whole of $X$, i. e. to a $x^* \in X^*$ with $x^*|_Y = y^*$, with the same norm, i. e. $$ \|x^{ * } \| = \|y^{* }\|.$$
Note that not every extension is a Hahn-Banach-extension.
Let's compute $g$'s norm, for $y \in Y$ we have $|y_1| = |y_2|$, that is $\|y\| = |y_1|$, giving
$$ \|g\| = \sup_{\|y\| \le 1} |g(y)| = \sup_{\|y\| \le 1} |y_2| = 1 $$
An extension of $f$ of $g$ is uniquely determined by its value on $(1,1)$, as for $x \in \mathbb R^2$
\begin{align*}
f(x_1, x_2) &= f\left(\frac{x_1 + x_2}2 + \frac{x_1 - x_2}2, \frac{x_1 + x_2}2 - \frac{x_1 - x_2}2\right)\\
&= f\left(\frac{x_1 + x_2}2, \frac{x_1 + x_2}2\right) + f\left(\frac{x_1 - x_2}2, -\frac{x_1 - x_2}2\right)\\
&= \frac{x_1 + x_2}2\cdot f(1,1) + g\left(\frac{x_1 - x_2}2, -\frac{x_1 - x_2}2\right)\\
&= \frac{x_1 + x_2}2 \cdot f(1,1) -\frac{x_1 - x_2}2\\
&= \left(\frac 12 f(1,1) + \frac 12\right)x_1 + \left(\frac 12 f(1,1)- \frac 12\right)x_2
\end{align*}
So we have $\def\abs#1{\left|#1\right|}\def\norm#1{\left\|#1\right\|}$
\begin{align*}
\|f\| &= \frac 12 \abs{f(1,1) + 1} + \frac 12\abs{f(1,1) - 1}\\
&\ge \frac 12 \abs{f(1,1) + 1 - f(1,1) + 1}\\
&= 1
\end{align*}
by the triangle inequality. We have equality iff $f(1,1) + 1$ and $1-f(1,1)$ have the same sign, that is iff $|f(1,1)| \le 1$. So any $f_\alpha$,
$$ f_\alpha(x) = \frac{x_1}2(\alpha + 1) + \frac{x_2}2(\alpha - 1), \qquad \alpha \in [-1,1] $$
is a H-B-extension. So 1., 2. are wrong and 3. is true.
