= Solution
For a bounded function on an interval $I$, write $\omega_I(f)=\sup_I f-\inf_I f$ for its <oscillation of a function on an interval>. For a partition $P$, the difference between its <upper Darboux sum> and <lower Darboux sum> is $\sum_{I\in P}|I|\omega_I(f)$. The <Riemann integrability criterion> says that this can be made arbitrarily small.
Since $\omega_I(f_1+f_2)\le\omega_I(f_1)+\omega_I(f_2)$, choose good partitions for the two functions and take their common refinement. <Darboux sum refinement monotonicity> makes both oscillation sums small on this refinement, so their sum is <Riemann integrable>. The same argument handles differences and scalar multiples.
Both $t\mapsto\max(t,0)$ and $t\mapsto|t|$ are <Lipschitz continuous> with constant one. Thus $\omega_I(f^+)\le\omega_I(f)$ and $\omega_I(|f|)\le\omega_I(f)$. This calculation shows that <Lipschitz composition preserves Riemann integrability>, so the <positive part> and the <absolute value> are integrable.
The converse for the absolute value is false. Define $f=1$ on <rational numbers> and $f=-1$ on <irrational numbers>. Then $|f|=1$ is integrable, but every interval has supremum one and infimum minus one, so every Darboux difference for $f$ is $2(b-a)$. Hence $f$ is not <Riemann integrable>.
Finally,
$$
\boxed{\max(f_1,f_2)=\frac{f_1+f_2+|f_1-f_2|}{2}}
$$
is integrable by the closure properties just proved. This establishes that <Riemann integrability is closed under addition and maximum>.
A countable supremum can fail, even for a uniformly bounded family. Enumerate the rational points of $[a,b]$ as $q_1,q_2,\ldots$ and set $f_n=1_{\{q_n\}}$. Each singleton <indicator function> is <Riemann integrable> with integral zero: isolate its single nonzero point in intervals of arbitrarily small total length. But
$$
\boxed{\sup_n f_n=1_{\mathbb Q\cap[a,b]}}
$$
is the nonintegrable <Dirichlet function>. This proves that a <pointwise supremum need not preserve Riemann integrability>.
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