= Solution
For the sum $F=f+g$:
* In case (a), \b[yes]. If $F$ were <differentiable function>[differentiable] at $a$, then $g=F-f$ would be differentiable there, contrary to the hypothesis.
* In case (b), \b[no]. Take
$$
f(x)=|x-a|,\qquad g(x)=-|x-a|.
$$
Both are continuous and nondifferentiable at $a$, but $F=0$ is differentiable.
For the product $G=fg$, the answer is \b[no] in both cases:
* For case (a), take $f(x)=x-a$ and $g(x)=|x-a|$. Then $f$ is differentiable at $a$, $g$ is not, but
$$
G(x)=(x-a)|x-a|
$$
has derivative $0$ at $a$.
* For case (b), take $f(x)=g(x)=|x-a|$. Neither factor is differentiable at $a$, whereas
$$
G(x)=(x-a)^2
$$
is differentiable.
Every displayed $f$ and $g$ is a <continuous function> and is nonzero at points arbitrarily close to $a$, so the extra condition in the question is satisfied.
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