$f,g:[0,+\infty) \to \mathbb R$ are real functions with Laplace transform $F$ and $G$.
So we can guarantee that:
$\square$: $(\cos t) g$ has Laplace transform $\dfrac {Gs}{s^2 +1}$
$\square$: $(\sin t) f$ has Laplace transform $\dfrac {F}{s^2 +1}$
$\square$: $c_1 f + c_2 g$ has Laplace transform $c_1 F + c_2 G$ for $c_1,c_2 \in \mathbb R$
$\square$: $cfg$ has a Laplace transform and $F G$, to $c \in \mathbb R$ [sic]
I have this question about the properties of the laplace transform and I'm not sure about the answer.
The third option is true according to the property of the linearity of the laplace transform. The others seem to be false. Am i missing something?
The fourth option has bad syntax and doesn't make sense as written. I imagine it's asking if $\mathcal L[cfg] = c FG$ is true for $c \in \mathbb R$.
You're correct that the $\mathcal L[f\cdot g] \ne \mathcal L[f]\cdot\mathcal L[g]$ in general. The correct relationship between the product of functions and the Laplace transform is given by the convolution theorem:
$\mathcal L[f*g] = \mathcal L[f]\cdot\mathcal L[g]$
The definition of $*$ is:
$(f * g)(t) = \displaystyle \int_0^t f(s)g(t-s)\,\mathrm ds$
(edit: In some contexts, the definition of $*$ has the bounds of integration be $-\infty$ to $+\infty$)
This is pronounced "$f$ convolved with $g$" or "the convolution of $f$ with $g$"
If you're asked to prove the first and second answers wrong, it's enough to find one counterexample.
Khan Academy has a few videos on this.
https://www.khanacademy.org/math/differential-equations/laplace-transform/convolution-integral/v/introduction-to-the-convolution