In general, it's possible for a composition to be invertible and neither function to be invertible: take $f\colon\{a\}\to\{1,2\}$ given by $f(a)=1$, and $g\colon\{1,2\}\to\{x\}$ given by $g(1)=g(2)=x$. Neither $f$ nor $g$ are invertible ($f$ is not onto, $g$ is not one-to-one) but $gf\colon\{a\}\to\{x\}$ is clearly invertible.
What is true is that if $f$ and $g$ are invertible and you can compose them, then $gf$ is invertible and $(gf)^{-1} = f^{-1}g^{-1}$. But you cannot conclude that $f$ and $g$ are each invertible from the assumption that $gf$ is invertible.
For a linear example, take $T\colon\mathbb{R}^2\to\mathbb{R}^3$ given by $T(x,y) = (x,y,x-y)$ (not onto), and $S\colon\mathbb{R}^3\to\mathbb{R}^2$ given by $S(x,y,z) = (x,y)$ (not one-to-one). Neither $T$ nor $S$ are invertible, but $ST\colon\mathbb{R}^2\to\mathbb{R}^2$ is invertible.
For a linear example with maps from the a vector space to itself, necessarily infinite dimensional in view of the theorem above, take the vector space of all sequences of real numbers, and let $T$ be the "right shift" operator and $S$ the "left shift" operator. that is, $T(a_1,a_2,a_3,\ldots) = (0,a_1,a_2,\ldots)$, and $S(a_1,a_2,a_3,\ldots) = (a_2,a_3,\ldots)$. Check that neither $T$ nor $S$ are invertible, but that $ST$ is the identity map, hence invertible.
So: first assume that $ST$ is invertible. Use this to show that $T$ must be one-to-one.
Added hint: To show that $T$ is one-to-one, you want to show that if $\mathbf{v}\in V$ is such that $T(\mathbf{v})=\mathbf{0}$, then $\mathbf{v}=\mathbf{0}$ (that is, the only vector that goes to $\mathbf{0}$ is the zero vector; we're showing the nullspace of $T$ is trivial, or that the nullity is $0$). So, let $\mathbf{v}$ be a vector in $V$ such that $T(\mathbf{v})=\mathbf{0}$. What is $ST(\mathbf{v})$? What does that tell you about $\mathbf{v}$? What does that tell you about $T$ (and why)?
Once you know that $T$ is one-to-one, then you can use the Rank-Nullity Theorem to conclude that since $T$ is one-to-one and maps from $V$ to itself ($V$ finite dimensional), then $T$ is invertible. And then you can use that both $ST$ and $T$ are invertible (and hence both $ST$ and $T^{-1}$ are invertible) to show that $S$ is invertible. And if both $S$ and $T$ are invertible, then show that $TS$ is invertible. This proves that if $ST$ is invertible, then $TS$ is invertible; the converse follows by simply swapping the roles of $S$ and $T$.
Best Answer
Neither of the options are necessarily true.
$T_1$ and $T_2$ can both be invertible, for example if $n=m$ and $T_1$ and $T_2$ are both the identity map.
On the other hand, both $T_1$ and $T_2$ can fail to be invertible, for example if $n=1$, $m=2$ and $T_1(x)=(x,0)$ and $T_2(x)=(x,0)$.