= Solution
For a <self-adjoint C-star element> $a$, the following are three equivalent characterizations of a <Positive element of a C-star algebra>:
$$
\boxed{\begin{aligned}
&\sigma(a)\subseteq[0,\infty);\\
&\|a-\|a\|1_A\|\leq\|a\|;\\
&\text{there is }t>0\text{ with }\|a-t1_A\|\leq t.
\end{aligned}}
$$
If $a=b^2$ with $b=b^*$, the <spectral mapping theorem> gives nonnegative spectrum. Conversely, the <continuous functional calculus> applied to $\sqrt{\lambda}$ on $\sigma(a)$ gives a <self-adjoint C-star element> $b$ with $b^2=a$. For real $\sigma(a)$,
$$
\|a-t1_A\|=\max_{\lambda\in\sigma(a)}|\lambda-t|.
$$
Thus the third condition means $\sigma(a)\subseteq[0,2t]$, while the first implies the second by taking the interval $[0,\|a\|]$. The second implies the third with $t=\|a\|$ if $a\ne0$; if $a=0$, any $t>0$ works. This proves every equivalence.
The <positive cone of a C-star algebra> contains zero and is closed under multiplication by nonnegative real scalars. To verify addition without assuming that positive elements commute, put $r=\|a\|$, $s=\|b\|$ for $a,b\geq0$. Then
$$
\|(a+b)-(r+s)1_A\|\leq\|a-r1_A\|+\|b-s1_A\|\leq r+s.
$$
If $r+s>0$, the third characterization makes $a+b$ positive; otherwise both are zero. Hence it is a <convex cone>. If $a$ and $-a$ are positive, $\sigma(a)=\{0\}$ and $a=0$ by the <self-adjoint> spectral <norm> formula; therefore the cone is pointed and defines a <partial order> on $A_{\mathrm{sa}}$ by $h\leq k$ if $k-h\in A_+$. It is generating because the <continuous functional calculus> gives
$$
|h|=(h^2)^{1/2},\qquad h_\pm=\tfrac12(|h|\pm h)\geq0,\qquad h=h_+-h_-.
$$
For $a\geq0$, the <spectral mapping theorem> yields
$$
\boxed{\|a\|\leq1\iff\sigma(a)\subseteq[0,1]\iff1_A-a\geq0\iff0\leq a\leq1_A.}
$$
Finally, let $a\in M_2(\mathbb C)$ be <Hermitian>, and write $\tau=\operatorname{tr}a$, $d=\det a$. Its two <eigenvalues> are real. Both are nonnegative exactly when their sum $\tau$ and product $d$ are nonnegative. The corresponding sum and product for $I-a$ are $2-\tau$ and $1-\tau+d$. Therefore the necessary and sufficient conditions for a <positive contraction> are
$$
\boxed{0\leq\tau\leq2,\qquad d\geq0,\qquad1-\tau+d\geq0.}
$$
Indeed, nonnegative sum and product force two real numbers to be nonnegative, including the zero-product cases. Applying this to $a$ and $I-a$ places both <eigenvalues> in $[0,1]$.
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