I am studying about $C^{*}$-algebras, but I face several problems.
Let $A$ be a $C^{*}$-algebra with identity, $a \in A$, $a \geq 0$.
1: Please show me how I prove that $ a \leq 1_{A} \Longleftrightarrow \Vert a \Vert \leq 1 $
2: Let $A$ be a $C^{*}$-algebra with identity $ a,b \in A$, if $a \geq 0$, $b \leq 0 $, then $ \sigma ( ab ) \subset [ 0, \infty ) $ and $ ab \geq 0 $
If $0\leq a\leq 1_A$, then $1_A-a\geq0$. Thus $\sigma(a)\subset[0,\infty)$ and $$1-\sigma(a)=\sigma(1_A-a)\subset[0,\infty),$$ so $\sigma(a)\subset[0,1]$. As $a$ is selfadjoint, $\|a\|=\max\sigma(a)\leq1$.
Conversely, if $\|a\|\leq1$: as the spectral radius is always at most the norm, and since $a\geq0$, we get $\sigma(a)\subset[0,1]$. Then $\sigma(1_A-a)\subset[0,1]$, and so $1_A-a\geq0$.
For the second part, if $a\geq0$ and $b\leq0$, then the spectrum of $ab$ is not necessarily positive even when $A=\mathbb C$ (as pointed out by Aweygan). If $a\geq0$ and $b\geq0$, then one can use that in a Banach algebra $$\{0\}\cup\sigma(xy)=\{0\}\cup\sigma(yx).$$ Then $$ \{0\}\cup\sigma(ab)=\{0\}\cup\sigma(a^{1/2}ba^{1/2})\subset[0,\infty). $$ Finally, if $A$ is not commutative, it is not true in general that the product of positive operators is positive. This fails already in $M_2(\mathbb C)$.