4D (Analysis I)

For the sum

F = f + g

In case (a), yes. If $F$ were differentiable at $a$ , then $g = F - f$ would be differentiable there, contrary to the hypothesis.
In case (b), no. Take
$f (x) = ∣ x - a ∣, g (x) = - ∣ x - a ∣.$
(1)
Both are continuous and nondifferentiable at $a$ , but $F = 0$ is differentiable.

For the product

G = f g

, the answer is 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
$G (x) = (x - a) ∣ x - a ∣$
(2)
has derivative $0$ at $a$ .
For case (b), take $f (x) = g (x) = ∣ x - a ∣$ . Neither factor is differentiable at $a$ , whereas
$G (x) = (x - a)^{2}$
(3)
is differentiable.