With $n \geq 2$, $X_n$ is a r.v. such that $$P\left(X_n = \frac{1}{n}\right) = 1 - \left(\frac{1}{n^2}\right)$$ and $$P(X_n = n) = \frac{1}{n^2}$$
I got that $\lim_{n \to \infty} E[X_n] = 0$ since in limits $P\left(X_n = \frac{1}{n}\right) = 1 - \left(\frac{1}{n^2}\right)$ converges to 1. Giving me $X_n = 1/n$ with probability 1.
But I am stuck with $\lim_{n \to \infty}\operatorname{Va}[X_n]$
I used the general variance formula: $$\operatorname{Var}(X) = E[X^2] - E[X]^2$$
Inserting $\frac{1}{n}$ for X: $$\operatorname{Var}(X) = \frac{1}{n^2} - \left(\frac{1}{n}\right)^2 = 0$$ Taking the limit also yields $0$, which seems incorrect.
"Inserting $\frac{1}{n}$ for $X$" is wrong (and so is your expression for $\mathbb{E}[X_n]$). First, $$\mathbb{E}[X_n]=n\cdot \frac{1}{n^2}+ \frac{1}{n}\cdot\left(1-\frac{1}{n^2}\right) = \frac{2}{n}-\frac{1}{n^3}$$ Second, $$\mathbb{E}[X_n^2]=n^2\cdot \frac{1}{n^2}+ \frac{1}{n^2}\cdot\left(1-\frac{1}{n^2}\right) =1+\frac{1}{n^2}-\frac{1}{n^4}$$
Therefore, $$\mathbb{E}[X_n] = \xrightarrow[n\to\infty]{} 0$$ and $$\operatorname{Var} X_n = 1+\frac{1}{n^2}-\frac{1}{n^4} - \left( \frac{2}{n}-\frac{1}{n^3}\right)^2\xrightarrow[n\to\infty]{} 1$$
Important conceptual mistake: You do not have "$X_n=1/n$ with probability 1." You can't take the limit in half the expression (the probability "$1-1/n^2 \xrightarrow[n\to\infty]{} 1$") but not for the rest (still having some $n$'s floating around).