1. We will call a threat an uncolored segment whose coloring of red would produce a triangle. We will show that Alice can use the following strategy to win: Alice will keep coloring segments whose one end-point is $A$ unless she can complete a triangle. We will call the other endpoints opposite ends. Let us assume that Bob can stop this strategy. Bob's optimal strategy to counter Alice's is first eliminating all threats in his move and then using his remaining moves to color segments coming out of $A$. Note that after the $k$-th red line is drawn, Bob must color $k - 1$ threats blue, which are the lines connecting the new opposite end with all the previous opposite ends. Thus in the $k$-th pair of moves of Alice and Bob, the number of colored edges coming out of $A$ has increased by $b + 1 - (k - 1) = b - k + 2$. Assume there exists a $l$ such that after $l$ moves all segments out of $A$ have been colored before Alice could draw a triangle. Then we have
$$(b + 2)l - \frac{l(l + 1)}{2} \geq n - 1 \Rightarrow l^2 + 2l\left(b + \frac{3}{2}\right) + 2n - 2 \leq 0,$$
therefore the discriminant of the quadratic equation is non-negative, i.e. $4\left(b + \frac{3}{2}\right)^2 - 4(2n - 2) \geq 0$ and therefore $b \geq \sqrt{2n - 2} - \frac{3}{2}$. Since this inequality is violated, it follows that Alice's winning strategy works.

2. We will show that Bob can use the following strategy to win: Whenever Alice colors a line $AB$, Bob will color $\lfloor b/2 \rfloor$ edges out of $A$ blue and $\lfloor b/2 \rfloor$ edges out of $B$. This ensures at most $n/\lfloor b/2 \rfloor + 1$ red edges out of any point, which is not larger than $\lfloor b/2 \rfloor + 1$ for $b \ge 2\sqrt{n}$. Assuming Alice completes a triangle $ABC$ and WLOG moves $AB$, $AC$ and $BC$ in sequence, it follows that before placing $BC$ there were at least $\lfloor b/2 \rfloor + 1$ threats out of $C$. However this means, since placing $AC$ created these threats, that, not counting $AC$, there were at least $\lfloor b/2 \rfloor + 1$ red lines out of $A$ and thus the total number of red lines coming out of $A$ is larger than $\lfloor b/2 \rfloor + 1$, a contradiction.

$\Box$