a) Actually, if $f(a)\neq c$, then $f(a)>c$ simply because the counterdomain of $f$ is $[c,b]$, not because $f$ is monotonic. Also, there is no need for $x\neq c$. You're correct in implying that $f(x)<f(a)$ and thus $x<a$, which is a contradiction.
b) I'm going to do some more details: Let $x\in[a,b]$. For simplicity, suppose that $x\neq a$, $x\neq b$ (otherwise, you have to check only left or right continuity, which is very similar to this case). Let $\varepsilon>0$, and consider $\varepsilon'=\frac{1}{2}\min(d-f(x),f(x)-c,\varepsilon)$ (notice that $\varepsilon'$ depends only on $\varepsilon$), so that $\varepsilon'>0$ and the interval $(f(x)-\varepsilon',f(x)+\varepsilon')$ is contained in $(c,d)$. By surjectivity, there exists $x_{\varepsilon}^-,x_{\varepsilon}^+\in[a,b]$ such that $f(x_{\varepsilon}^\pm)=f(x)\pm\varepsilon'$. Since $f$ is monotonic, then $x_{\varepsilon}^-<x<x_{\varepsilon}^+$.
Let $\delta=\frac{1}{2}\min(x_{\varepsilon}^+-x,x-x_{\varepsilon}^-)$ (notice that $\delta$ depends solely on $\varepsilon$). Given any $y\in(x-\delta,x+\delta)$, we have $x_{\varepsilon}^-<x-\delta<y<x+\delta<x_{\varepsilon}^+$, so, since $f$ is monotonic,
$$f(x)-\varepsilon<f(x)-\varepsilon'=f(x_\varepsilon^-)<f(y)<f(x_\varepsilon^+)=f(x)+\varepsilon'<f(x)+\varepsilon$$
because $\varepsilon'<\varepsilon$. Thus, we have just shown that $f(y)\in(f(x)-\varepsilon,f(x)+\varepsilon)$ whenever $y\in(x-\delta,x+\delta)$, so $f$ is continuous at $x$.
Since $x$ was arbitrary (and assuming you have already shown the cases $x=a$ and $x=b$), then $f$ is continuous.
c)By assumption, $f$ is surjective and, since $f$ is monotonic, it is injective, so there exists an inverse $f^{-1}$. This inverse is also surjective and, you can easily check, monotonic. By item b), with $f^{-1}$ in place of $f$, $f^{-1}$ is continuous.
To prove statement A, let $x > y$. Then, as we know, $f(x) > f(y)>0$ and $g(x) > g(y)>0$. Hence, the following chain of statements proves the claim:
\begin{gather}
f(x) > f(y) \implies g(x)f(x)> g(x)f(y) \quad(\because g(x)>0)\\
g(x) > g(y) \implies g(x)f(y) > g(y)f(y) \quad(\because f(y)>0)\\
g(x)f(x)> g(x)f(y) > g(y)f(y) \implies fg(x) > fg(y)
\end{gather}
An analogous proof would follow for part B if $f$ and $g$ were increasing, with a caveat:let $x > y$. Then, as we know, $f(x) > f(y)$ and $g(x) > g(y)$. Hence, the following chain of statements proves the claim:
\begin{gather}
f(x) > f(y) \implies g(x)f(x)< g(x)f(y) \quad(\because g(x)<0)\\
g(x) > g(y) \implies g(x)f(y) < g(y)f(y) \quad(\because f(y)<0)\\
g(x)f(x)< g(x)f(y) < g(y)f(y) \implies fg(x) < fg(y)
\end{gather}
Now, let us see if the same logic could work with part A':let $x > y$. Then, as we know, $f(x) < f(y)$ and $g(x) < g(y)$. Hence, the following chain of statements proves the claim:
\begin{gather}
f(x) < f(y) \implies g(x)f(x)< g(x)f(y) \quad(\because g(x)>0)\\
g(x) < g(y) \implies g(x)f(y) < g(y)f(y) \quad(\because f(y)>0)\\
g(x)f(x)< g(x)f(y) < g(y)f(y) \implies fg(x) < fg(y)
\end{gather}
That's brilliant, so great intuition for anticipating part A'. Now we will check part B':let $x > y$. Then, as we know, $f(x) < f(y)$ and $g(x) < g(y)$. Hence, the following chain of statements proves the claim:
\begin{gather}
f(x) < f(y) \implies g(x)f(x)> g(x)f(y) \quad(\because g(x)<0)\\
g(x) < g(y) \implies g(x)f(y) > g(y)f(y) \quad(\because f(y)<0)\\
g(x)f(x)> g(x)f(y) > g(y)f(y) \implies fg(x) > fg(y)
\end{gather}
And therefore part B' is also done. Note the above logic carefully, I think all steps are equally important.
Use this logic, and see why in most cases, one increasing and one decreasing function doesn't tell you much about the product itself.
Best Answer
This is not true. Consider $f(x)=e^x$ and $g(x)=-e^x$. Then $(fg)(x)=-e^{2x}$ is strictly decreasing.