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Let $X_n$ be a random variable sequence defined as below

$$
P(X_n=x_n)=
begincases
1-left(dfrac12right)^n&textfor x_n=0 \
left(dfrac12right)^n&textfor x_n=1
endcases
$$
for $n=1,2,3,ldots$.

(a) Show that $X_n$ almost sure converges.
(b) Show that $X_n$ mean square converges.

(a) $$
limlimits_ntoinfty P(X_n=0) =limlimits_ntoinftyleft( 1-left(dfrac12right)^nright) = 1-0 = 1.
$$
$$
limlimits_ntoinfty P(X_n=1) =limlimits_ntoinftyleft(dfrac12right)^n= 0.
$$

So, I conclude that $X_n=0$ is almost sure convergent, but $X_n=1$ is not almost sure convergent.

Is it true? Please check my answer.

(b)
$$
limlimits_ntoinfty Eleft((X_n-X)^2right) = limlimits_ntoinfty Eleft(X_n^2-2X_nX+X^2right)
=
limlimits_ntoinfty Eleft(X_n^2)-2E(X_nX)+E(X^2)right).
$$
I know how to check mean square convergence, is
$$limlimits_ntoinfty Eleft((X_n-X)^2right)=0.$$
Now, I confuse about $E(X)$. How to find $E(X)$?

I guess you are a beginner in this topic, so I'll write my answer as clearly as possible. You must have missed something during your study about the convergence of random variables. The terms in my answer are based on this Wikipieda page.

For (a), what you proved is that $X_n$ converges to $0$ in probability. But your "conclusion" I conclude that $X_n=0$ is almost sure convergent, but $X_n=1$ is not almost sure convergent does not make any sense. Think carefully about that.

Now I move to the proof of your problem.

(a) Since
$$sum_n=1^infty P(|X_n|>0)=sum_n=1^infty left(frac12right)^n<infty,$$
we know that $P(limsup_ntoinfty)=0$ by Borel-Cantelli lemma. This implies that $P(textevents text take place only for finitely many times)=1,$ which means that $X_n$ converges to $0$ almost surely.

(b) In this case $X=0$ so $E((X_n-X)^2)=E(X_n^2)=(frac12)^nto 0$. Therefore $X_n$ converges to $0$ in mean square.

You cannot prove almost sure convergence by just showing that $P(X_n=0) to 1$ and $P(X_n=1) to 0$. Note that $sum_m P(X_n neq 0)=sum_n frac 1 2^n <infty$. By Borel Cantelli Lemma this implies that with probability $1$ the sequence $(X_n)$ becomes $0$ after some stage. Hence $X_n to 0$ almost surely.

Also $EX_n^2=frac 1 2^n to 0$ so $X_n to 0$ in mean square.