some theorens in sequences I need hint to solve 2 and 3 and in 1 is my answer right? .
True or false with justification:
1) $\displaystyle X_n \to L \implies \frac {X_1+X_2+\cdots+X_n}{n}\to L $
True,
proof: 
Claim,
$$\lim_{n\to\infty} \left(\frac{X_1+X_2+\cdots+X_n}{n}-L\right)=0 $$
$$=\lim_{n\to\infty} \left(\frac{X_1+X_2+\cdots+X_n-nL} n \right) = \lim_{n\to\infty} \frac{(X_1-L) + \cdots+(X_n-L)}{n}$$
but, 
$\mathcal X_n \to L$ 
i.e let $\epsilon$ be given then $\exists n_0 \text{ s.t. } \vert {X_n-L}|< \epsilon  \forall n>n_0 $
Now,
\begin{align}
& \left|\frac{(X_n-L)+\cdots+(X_{n0}-L)+(X_{n0+1})+\cdots+(X_n-L)}n \right| \\
\le {} & \left|\frac{X_1-L}n \right| + \cdots + \left|\frac{X_{n0}-L} n \right| + \left|\frac{X_{n0+1}}n \right| +\cdots+ \left|\frac{X_n-L}{n}\right| \\
\le {} & 0 +\frac \epsilon n + \cdots+\frac \epsilon n = \left(\frac{n-n_0} n \right) \epsilon\to \epsilon
\end{align}
2) $X_n \to 1 \implies\sqrt{X_n} \to 1$
3) $\frac{X_n-1}{X_n+1}\to 0 \implies X_n \to 1 $

How could you replace the first $n_0$ terms by $0$ and have that inequality? Your approach is in correct direction, but not correct.
Fix:
Since the sequence is convergent, it is bounded. Thus, $|x_i-L|<M$ for some $M$ independent of index $i$. Now write:
\begin{align}
& \left|\frac{(X_n-L)+\cdots+(X_{n0}-L)+(X_{n0+1})+\cdots+(X_n-L)}n \right| \\
\le {} & \left|\frac{X_1-L}n \right| + \cdots + \left|\frac{X_{n0}-L} n \right| + \left|\frac{X_{n0+1}}n \right| +\cdots+ \left|\frac{X_n-L}{n}\right| 
\le n_o\frac{M}{n} +\frac \epsilon n + \cdots+\frac \epsilon n = \frac{n_0M}{n}+\left(\frac{n-n_0} n \right) \epsilon\to \epsilon
\end{align}
So, given $\epsilon >0$ first find $n_0$ such that $|x_i-L| < \epsilon /2$ for all $i>n_0$, and then find a second $n_1 > n_0$ big enough so that $n_0m/n < \epsilon /2$ if $n>n_1$. Now, for any $n>n_1$ we'll have, from above, that
$$\frac{(X_n-L)+\cdots+(X_{n0}-L)+(X_{n0+1})+\cdots+(X_n-L)}n < \epsilon .$$
This proves 1).
