suppose that $A$ is a Banach algebra with identity, $a, b \in A$:
are these correct?
$ \sigma(a+b) \subseteq \sigma(a) +\sigma(b) $
$ \sigma(ab)\subseteq \sigma(a)\sigma(b) $
can you show me that abelian condition $A$ is necessary? please give me an example to show that the above conditions is not true, if the abelian condition does not exist.
No not in general. However, if you assume that $ab=ba=$, then they hold. To see this, assume at first $A$ is commutative. By considering the Banach algebra generated by the set $$\lbrace 1_A,a,b, a^*,b^*, (z_1 1_A - a)^{-1} , (z_2 1_B - b)^{-1}: z_1 \in \rho(a), z_2\in \rho(b) \rbrace.$$
this may be assumed, since this does not alter the spectra and defines a commutative Banach algebra with $1_A$ as a unit (here $\rho$ refers to the resolvent).
Let $\text{Prim}(A)$ denote the set of all maximal ideals in $A$ and let $\tilde{A}$ be the space of characters on $A$, so that $\tilde{A} \cong \text{Prim}(A)$. Pick some $\lambda\in \sigma(ab)$ so that $\lambda - ab$ is non-invertible. Being non-invertible, it generates a two-sided closed ideal $I$ differing from $A$, which must be contained in some maximal ideal $J$ (or be one itself). The maximal ideal $J$ corresponds to some character $\varphi \colon A\rightarrow \mathbb{C}$ according to the above isomorphism. Since $\lambda - ab$ is non-invertible in a commutative C$^\ast$-algebra, it follows that $\lambda - \varphi(ab) = \varphi(\lambda - ab) = 0$.
Now, as $\varphi$ is a character we have $\varphi(a)\varphi(b)=\lambda$ while $$ \varphi(\varphi(a)1_A - a) = 0 $$
shows that $\varphi(a)$ must be in the spectrum of $a$ with a similar observation giving $\varphi(b) \in \sigma(b)$ (recall that in a commutative Banach algebra, an element $x$ are invertible if and only if $\varphi(x) \neq 0$ for every character $\varphi$.). So $\lambda = \varphi(a)\varphi(b) \in \sigma(a)\sigma(b)$ as desired.