All are true.
A:
Suppose $g \circ f(x)=g \circ f(y)$, i.e., $g(f(x))=g(f(y))$. Since $g$ is one-to-one, then $f(x)=f(y)$. Since $f$ is one-to-one, then $x=y$. Therefore $g \circ f$ is one-to-one.
B:
Let $\mathrm{c} \in \mathrm{C}$. Since $\mathrm{g} \circ \mathrm{f}$ is onto, there exists an $\mathrm{a} \in \mathrm{A}$ such that $\mathrm{g} \circ \mathrm{f}(\mathrm{a})=\mathrm{c}$.
Let $b=f(a) \in B$. Then $g(b)=g(f(a))=c$. So $g(b)=c$. Therefore $g$ is onto.
C:
Suppose $f(x)=f(y)$, then $g(f(x))=g(f(y))$. Since $g \circ f$ is one-to-one and $g \circ f(x)=g \circ f(y)$, then $x=y$. Therefore $f$ is one-to-one.
An alternative proof would be to assume for the sake of contradiction that $\mathrm{f}$ is not one-to-one, i.e., there exist distinct $\mathrm{x}$ and $y$ such that $\mathrm{f}(\mathrm{x})=$ $\mathrm{f}(\mathrm{y})$. But then $g(\mathrm{f}(\mathrm{x}))=\mathrm{g}(\mathrm{f}(\mathrm{y})$, which implies that $\mathrm{x}=\mathrm{y}$, contradicting the fact that $\mathrm{x}$ and $\mathrm{y}$ are distinct.
The two proofs are very similar but I wrote both of them to illustrate that you don't have to think about it a certain way.
D:
Let $x, y \in B$ and suppose $g(x)=g(y)$. Since $x \in B$ and $y \in B$ and $f$ is onto, there exist $a_{1}, a_{2} \in A$ such that $f\left(a_{1}\right)=x$ and $f\left(a_{2}\right)=y$. Therefore $g\left(f\left(a_{1}\right)\right)=g\left(f\left(a_{2}\right)\right)$, so $g \circ f\left(a_{1}\right)=g \circ f\left(a_{2}\right)$. Since $g \circ f$ is one-to-one, then $a_{1}=a_{2}$. But then $f\left(a_{1}\right)=f\left(a_{2}\right)$, so $x=y$. Therefore $g$ is one-to-one.
If $g \circ f$ is onto and $g$ is one-to-one, show that $f$ is onto.
Let $b \in B$. Now consider $c=g(b) \in C$. Since $g \circ f$ is onto, then there exists an $a \in A$ such that $g \circ f(a)=c$. Therefore $g(f(a))=c=g(b)$. Since $g$ is one-to-one, $f(a)=b$. So we've shown that there exists an $a \in A$ such that $f(a)=b$, which shows that $f$ is onto.
