Let $R = c (b-a)$ and let $\varepsilon > 0$.
You want to show that there is $\delta > 0$ such that for all tagged partitions $P$ with $\|P\| < \delta$ you have
$$ \left | \sum_{k=1}^n f(x_i) (t_{i+1}-t_i) - R \right | < \varepsilon$$
where $x_i \in [t_i , t_{i+1}] \subset [a,b]$ form a tagged partition of $[a,b]$. We have $f(x_i) = c$ hence
$$ \left | \sum_{k=1}^n f(x_i) (t_{i+1}-t_i) - R \right | = \left |c \sum_{k=1}^n (t_{i+1}-t_i) - R \right | = \left | c (b-a) - R \right | = 0 < \varepsilon$$
hence $f$ is Riemann integrable with $$ \int_a^b f(x) dx = c (b - a)$$
Hope this helps.
Consider a partition $a = x_0 < x_1 < \ldots < x_n = b$ and for a subinterval $I_j = [x_{j-1},x_j]$ define
$$D_{j}(g) = \sup_{x \in I_j}g(x) - \inf_{x \in I_j}g(x) = \sup_{x,y \in I_j}|g(x) - g(y)| \\ D_{j}(f \circ g ) = \sup_{x \in I_j}f(g(x)) - \inf_{x \in I_j}f(g(x)) = \sup_{x,y \in I_j}|f(g(x)) - f(g(y))|$$
Since $f$ is continuous it is uniformly continuous and bounded (by extension if necessary) on a closed interval $[c,d]$ such that $g([a,b]) \subset [c,d].$
Hence, $|f(x)| \leqslant M$ for $x \in [c,d]$ and for every $\epsilon >0$ there exists $\delta > 0$ such that if $|x_1 - x_2| < \delta$ then $|f(x_1) - f(x_2)| < \epsilon/(2(b-a))$ .
Since $g$ is integrable, if the partition norm $\|P\|$ is sufficiently small we have
$$U(P,g) - L(P,g) = \sum_{j=1}^n D_j(g) (x_j - x_{j-1}) < \frac{ \delta \epsilon}{4M}.$$
We can split the upper-lower sum difference $U(P,f \circ g) - L(P, f \circ g)$ into two sums as given by
$$\tag{1}U(P,f \circ g) - L(P, f \circ g) = \sum_{D_j(g) \geqslant \delta} D_j(f \circ g)(x_j - x_{j-1}) + \sum_{D_j(g) < \delta} D_j(f \circ g)(x_j - x_{j-1})$$
In the second sum on the RHS of (1) we have $D_j(f \circ g) < \epsilon/(2(b-a)$ since by uniform continuity $D_j(g) < \delta \implies |g(x) - g(y)| < \delta \implies |f(g(x)) - f(g(y))| < \epsilon/(2(b-a)$ for all $x,y \in I_j$.
Thus,
$$\tag{2}\sum_{D_j(g) < \delta} D_j(f \circ g)(x_j - x_{j-1}) < \frac{\epsilon}{2}$$
Considering the first sum on the RHS of (1), first note that
$$\sum_{D_j(g) \geqslant \delta} (x_j - x_{j-1}) \\ < \delta^{-1}\sum_{D_j(g) \geqslant \delta} D_j(g)(x_j - x_{j-1}) < \delta^{-1} [U(P,g) - L(P,g)] < \delta^{-1} \frac{\delta \epsilon}{4M} = \frac{\epsilon}{4M} .$$
Hence,
$$\tag{3}\sum_{D_j(g) \geqslant \delta} D_j(f \circ g)(x_j - x_{j-1}) < \sum_{D_j(g) \geqslant \delta} 2M(x_j - x_{j-1}) < \frac{\epsilon}{2}.$$
From (1), (2) and (3) we obtain
$$U(P,f \circ g) - L(P, f \circ g) < \epsilon,$$
and conclude that $f \circ g$ is integrable.
Best Answer
Hint. To show that $|f|$ is integrable Riemann, it suffices to show that for every partition $P$ of $[a,b]$ $$ U(|f|,P)-L(|f|,P)\le U(f,P)-L(f,P), $$ which in order to prove it suffices to show that, for every $[s,t]\subset [a,b]$ $$ \sup_{x,y\in[c,d]} |f(x)|-|f(y)|= \sup_{x\in[c,d]}|f(x)|- \inf_{x\in[c,d]}|f(x)|\le \sup_{x\in[c,d]}f(x)- \inf_{x\in[c,d]}f(x)=\sup_{x,y\in[c,d]} f(x)-f(y), $$ which is a consequence of the fact that $$ \big| |f(x)|-|f(y)|\big|\le |f(x)-f(y)|. $$
For the second part of the question, just use the fact that $$ -|f(x)|\le f(x)\le |f(x)| $$ and hence $$ \int |f| \le \int f \le \int |f| $$ and finally $$ \Big|\int f\,\Big| \le \int|f| $$