Convergence by probability and transformation Question 1:
If $X_n$ converges in probability to $X$, can I apply the continuous mapping theorem to say that $\dfrac{X_n}{n}$ converges in probability to $\dfrac{X}{n}$?
Continuous mapping theorem as per wikipedia: Suppose $g(\cdot)$ is a continuous function. If $ X_n $ converges in probability to $ X $, then $ g(X_n) $ converges in probability to $ g(X) $.
Question 2:
My metrics book stated that, if $X_n$ converges in probability to $X$, then $\dfrac{nX_n}{n - 3}$ will converge in probability to $X$. Why? Why is $\dfrac{n}{n - 3}$ consistent to $1$?
 A: I'm not sure how you can use it since $n$ also goes to infinity, but using the definition is also simple.
For $n\geq 1$, we have
$$\mathbb P(|X_n/n-X/n|>\epsilon)\leq \mathbb P(|X_n-X|>\epsilon)$$
because the LHS term is getting smaller when we divide it by $n$. Taking the limit makes the right of the inequality $0$, which makes the left of the inequality $0$ as well.
For the second one, you can use the theorem. You have $$Z_n=\left[ X_n\ \ Y_n\right]\rightarrow [X\ \ Y]$$
where $Y_n=n/(n-3)\rightarrow 1 = Y$, so $g(Z_n)=X_nY_n\rightarrow XY=X$.
A: In your first question, continuous mapping theorem is not applicable because the resulting limit should not depend on $n$. Moreover, $g(\cdot)$ in CMT is a continuous function that also should not depend on $n$, otherwise you would get a sequence of functions.
As for Question 2, I agree with the answer above. I can only add that the mentioned theorem is often called Slutsky's theorem (or the corollary from it). You can have a look at this Wikipedia article.
