The projectivisation of a complex rank $2$ vector bundle is not a complex line bundle, but rather a $\mathbb{CP}^1$ bundle (just as the projectivisation of $\mathbb{C}^2$ is not $\mathbb{C}$, but rather $\mathbb{CP}^1$).
As I mentioned in the comments, $\frac{i}{2\pi}\operatorname{trace}F_A$ is a representative for $c_1(E)$ by Chern-Weil theory. Therefore
$$\deg(E) = \frac{i}{2\pi}\int_X\operatorname{trace}F_A = \int_Xc_1(E).$$
So formulae for the degree follow from formulae for the first Chern class.
For the tensor product, you can show that $c_1(E_1\otimes E_2) = \operatorname{rank}(E_1)c_1(E_2) + \operatorname{rank}(E_2)c_1(E_1)$ by using the splitting principle or the Chern character; see this answer for both methods in the special case where one of the vector bundles is a line bundle.
As for the other two cases, recall that the first Chern class is additive in short exact sequences. From the short exact sequence
$$0 \to E_1 \to E_1\oplus E_2 \to E_2,$$
we see that $c_1(E_1\oplus E_2) = c_1(E_1) + c_1(E_2)$. Likewise, from the short exact sequence
$$0 \to F \to E \to E/F \to 0,$$
we see that $c_1(E) = c_1(F) + c_1(E/F)$, so $c_1(E/F) = c_1(E) - c_1(F)$. Note that this can also be deduced from the first short exact sequence because Chern classes are topological and short exact sequences of continuous vector bundles always split.
Therefore
\begin{align*}
\deg(E_1\otimes E_2) &= \operatorname{rank}(E_1)\deg(E_2) + \operatorname{rank}(E_2)\deg(E_1)\\
\deg(E_1\oplus E_2) &= \deg(E_1) + \deg(E_2)\\
\deg(E/F) &= \deg(E) - \deg(F).
\end{align*}
Note that these identities are true on a compact Kähler manifold too.
Best Answer
Suppose we have
$$0 \to \mathcal{O}(d) \to \mathcal{O}(a)\oplus\mathcal{O}(b) \to \mathcal{O}(c)\to 0$$
for some $c \in \mathbb{Z}$. Tensoring through by $\mathcal{O}(-d)$ gives the short exact sequence
$$0 \to \mathcal{O} \to \mathcal{O}(a - d)\oplus\mathcal{O}(b - d) \to \mathcal{O}(c-d) \to 0.$$
Therefore the bundle $\mathcal{O}(a - d)\oplus\mathcal{O}(b - d)$ has a non-zero section, so either $\mathcal{O}(a - d)$ has a section (in which case $a - d \geq 0$), or $\mathcal{O}(b - d)$ has a section (in which case $b - d \geq 0$). That is, $a \geq d$ or $b \geq d$, so $d \leq \max\{a, b\}$ as pointed out by Sasha in the comments.