does convergence of $X_{n}$ imply convergence of $VarX_{n}$? Can we conclude that $VarX_{n}\rightarrow VarX$ if $X_{n}$ converges
to $X$ almost surely and $VarX_{n}<\infty$ and $VarX<\infty$? What
about if $X_{n}$ converges in mean square?

No. Consider for example $\Omega := [-1,1]$ endowed with the Lebesgue measure and  $$X_n(\omega) := \begin{cases} n & 0 \leq x \leq \frac{1}{n^2} \\ -n & - \frac{1}{n^2} \leq x < 0 \\ 0 & \text{otherwise} \end{cases}, \qquad \omega \in \Omega. $$ Then $\mathbb{E}X_n = 0$, $X_n \to X := 0$ almost surely and
$$\text{var} \, X_n = \mathbb{E}(X_n^2) = 2 \frac{n^2}{n^2} = 2$$
This shows that $\text{var} \, X_n$ does not converge to $\text{var} \, X=0$.
If $X_n$ converges in mean-square to a random variable $X$, then $X_n \to X$ in $L^2$ and this implies, by the triangle inequality, in particular the convergence of the second moments.
