= Solution
For an element $a$ of a complex unital <Banach algebra>, its <spectrum of an element> is
$$
\boxed{\sigma_A(a)=\{\lambda\in\mathbb C:\lambda1-a\text{ is not invertible in }A\}.}
$$
The <nonzero spectra of products in opposite orders> agree:
$$
\boxed{\sigma(ab)\setminus\{0\}=\sigma(ba)\setminus\{0\}.}
$$
For $\lambda\ne0$, if $R=(\lambda1-ab)^{-1}$ exists, direct multiplication on both sides verifies
$$
(\lambda1-ba)^{-1}=\lambda^{-1}(1+bRa).
$$
Interchanging $a,b$ proves the converse. Zero may differ. On the <Hilbert space> $\ell^2(\mathbb N_0)$, let $S$ be the <unilateral shift operator> $Se_n=e_{n+1}$ and take $a=S^*$, $b=S$. Then $ab=S^*S=I$, whereas $ba=SS^*=I-P_0$, where $P_0$ is projection onto the first coordinate. Hence
$$
\boxed{\sigma(ab)=\{1\},\qquad\sigma(ba)=\{0,1\}.}
$$
Both values for the second operator occur on its two nonzero invariant summands; every other value has an inverse obtained separately on those summands.
We use the spectral definition of a <Positive element of a C-star algebra>: $h$ is positive when $h=h^*$ and $\sigma(h)\subseteq[0,\infty)$. The <continuous functional calculus>, supplied by the <Commutative Gelfand--Naimark theorem> for the algebra generated by a <Normal element of a C-star algebra> and its adjoint, sends $F\in C(\sigma(h))$ to $F(h)$. It preserves products, sums, conjugation and the identity, is isometric, and satisfies $\sigma(F(h))=F(\sigma(h))$. For a self-adjoint element the spectrum is real, so real-valued functions produce self-adjoint elements.
For example, $F(t)=\sqrt t$ gives the <positive square root in a C-star algebra> of a <Positive element of a C-star algebra>. It satisfies $(h^{1/2})^2=h$. Its uniqueness follows from the composition rule: if $d\geq0$ and $d^2=h$, then the <continuous functional calculus> in the algebra generated by $d$ gives $\sqrt{d^2}=d$. For positive invertible $h$, the functions $t^{-1}$ and $t^{-1/2}$ give positive inverses and inverse square roots. If $0\leq h$ and $r\geq\|h\|$, the function $r-t$ gives $r1-h\geq0$; the same calculus gives
$$
\|r1-h\|=\max_{t\in\sigma(h)}|r-t|\leq r.
$$
These transfer elementary scalar inequalities for functions of one element to the algebra. They do not justify applying a scalar inequality to arbitrary noncommuting elements.
To prove closure under addition without assuming an order theorem, let $a,b\geq0$, set $\alpha=\|a\|$, $\beta=\|b\|$ and $s=\alpha+\beta$. If $s=0$, both elements are zero. Otherwise,
$$
\|s1-(a+b)\|\leq\|\alpha1-a\|+\|\beta1-b\|\leq s.
$$
The element $a+b$ is self-adjoint, so any $\lambda\in\sigma(a+b)$ is real and satisfies $|s-\lambda|\leq s$. Thus $\lambda\geq0$, proving
$$
\boxed{a,b\geq0\Longrightarrow a+b\geq0.}
$$
We will also use that the positive cone is proper: if $h\geq0$ and $-h\geq0$, then $\sigma(h)\subseteq\{0\}$, and the norm formula for self-adjoint elements gives $h=0$.
We next prove $a^*a\geq0$ without presupposing that positivity is defined by such products. Put $h=a^*a$, which is self-adjoint, and form its negative part by the <continuous functional calculus>:
$$
h_- =F(h),\qquad F(t)=\max\{-t,0\}.
$$
This is positive. Let $c=ah_-^{1/2}$. Since $h_-$ commutes with $h$,
$$
c^*c=h_-^{1/2}hh_-^{1/2}=-h_-^2\leq0.
$$
The nonzero product-spectrum identity proved above gives $\sigma(cc^*)\setminus\{0\}=\sigma(c^*c)\setminus\{0\}$. As $cc^*$ is self-adjoint, it too has spectrum in $(-\infty,0]$, so $cc^*\leq0$.
Write $c=u+iv$ with $u=u^*$ and $v=v^*$. The <continuous functional calculus> makes $u^2$ and $v^2$ positive, and the addition result gives
$$
c^*c+cc^*=2(u^2+v^2)\geq0.
$$
But both summands on the left are nonpositive, so their sum is also nonpositive. Properness of the cone forces $c^*c+cc^*=0$. Now $c^*c=-cc^*$ is positive as well as nonpositive, so $c^*c=0$. The <C-star identity> gives $c=0$, whence $h_-^2=0$ and $h_-=0$. Thus $h$ has no negative spectral part:
$$
\boxed{a^*a\geq0\quad\hbox{for every }a\in A.}
$$
This proves the <positivity of adjoint products from spectral positivity> and the equivalence with the usual definition by elements of the form $d^*d$: the reverse direction is the factorization $h=(h^{1/2})^*h^{1/2}$ for $h\geq0$.
Finally, if $a\geq0$ and $c\in A$, write
$$
c^*ac=(a^{1/2}c)^*(a^{1/2}c).
$$
The preceding result therefore proves
$$
\boxed{c^*ac\geq0.}
$$
In particular, <congruence preserves positivity in a C-star algebra>, and $a\leq b$ implies $c^*ac\leq c^*bc$. These facts will justify the noncommutative order manipulations in the numbered parts.
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