Since the OP has indicated that he/she has solved it using my hints (see edit history and comments), I have edited to give a complete answer for future readers.
Statement a. While it is possible to take an arbitrary sum of tensors in $M'\otimes N'$ and show that it is in the image of $f\otimes g$, I will opt for using the universal property of tensor products and notion of epimorphisms as this will clarify the reason why Statement b is not true, let alone the dual of this one.
Let $h,k:M'\otimes N'\to L$ be arrows which satisfy $h\circ f\otimes g=k\circ f\otimes g$. Since the diagram below commutes note that we have $\bar h\circ f\times g=\bar k\circ f\times g$, where $\bar h:M'\times N'\to L$ is the composite of $h$ with the universal arrow. Since $f\times g$ is epi when both $f$ and $g$ are epi, $\bar h=\bar k$, and by the universal property of tensor products, $h=k$.
M\times N @>f\times g>> M'\times N' \\
@VVV @VVV \\
M\otimes N @>f\otimes g>> M'\otimes N'
Statement b. If you attempt to argue in the same way here, you come to a halt right at the start, and it becomes clear that there is no way to work with an equality like $f\otimes g\circ h=f\otimes g\circ k$. This gives a moral justification for the fact that this statement is false.
To construct a counterexample all we have to do is find a non-flat module $A$ and an injection which does not remain injective by tensoring with $A$. E.g. let $R=\mathbb Z$, $f:\mathbb Z\to\mathbb Z$ be multiplication by $2$, and $g:\mathbb Z/2\mathbb Z\to \mathbb Z/2\mathbb Z$ be the identity. Then $f\otimes g:\mathbb Z/2\mathbb Z\to\mathbb Z/2\mathbb Z$ is the zero map.