If $(a_{2k})$m $(a_{2k+1})$ and $(a_{3k})$ converge, then $(a_n)$ converge 
Work:
We may prove it by contradiction. If $a_n $ does not converge, then we can take some $\epsilon$ such that $\forall N $ and any $n> N$ one has $|a_n - L| \geq \epsilon$
Since this is true for any $N$ and any $n$, then it better be true for $n = 2k$ thus we see that $a_{2k}$ does not converge and we have reached a contradiction.
As for part (b), this is false. For example take $a_n = (-1)^n$ which diverges. However, $a_{2k} = 1$ converges to $1$ and $a_{2k+1} = -1$ converges to $-1$.
Is this correct?
 A: Consider the subsequence $$(b_n)=(a_{6n})$$
and
$$(c_n)=(a_{6n+3})$$
$(b_n)$ is a subsequence of
$(a_{2n})$ and of $(a_{3n})$.
$(c_n)$ is a subsequence of
$(a_{3n})$ and of $(a_{2n+1})$
thus
$(a_{2n})$ and $(a_{2n+1})$ converge to the same limite. We conclude that $(a_n)$ converges.
A: No, part (a) is not correct.
You have negated the definition of convergence incorrectly. Instead of 

... $\forall N$ and any $n>N$ one has $|a_n - L| \geq \epsilon$ 

it should be 

... $\forall N$ there exists $n>N$ such that $|a_n - L| \geq \epsilon$

Because of this, your statement "it better be true for $n=2k$" does not hold. In general, given $N$, the $n>N$ that you get might be in just one, or maybe two, or if you're lucky maybe all three of the given subsequences. So whatever contradiction you intend to find, it'll be subtler.
Another approach is the direct one. Assume that $a_{2k}$ converges to some limit, call that limit $L_1$. Assume also that $a_{2k+1}$ converges to some limit, call that limit $L_2$. And assume that $a_{3k}$ converges to some limit, call that limit $L_3$. 
The first thing to prove is that all these $L$'s are equal: first observe that $a_{2k}$ and $a_{3k}$ are both subsequences of $a_{6k}$, hence $a_{6k}$ converges to both $L_1$ and $L_3$, hence $L_1=L_3$. Similarly $a_{6k+3}$ is a subsequence of both $a_{2k+1}$ and $a_{3k}$, and so $L_2=L_3$.
Given $\epsilon>0$, apply the fact that $(a_{2k})$ converges to $L$ to find a $K_1$ such that if $k \ge K_1$ then $|a_{2k} - L| < \epsilon$. Similarly apply the fact that $(a_{2k+1})$ converges to some $L_2$ to find a $K_2$. Then take $N$ to be the maximum of $2K_1$, $2K_2+1$. 
If $n \ge N$ then $n$ is either even or odd so there exists some $k$ such that $2k=n$ or $2k+1=n$. In the first case $k \ge K_1$ so $|a_n-L|=|a_{2k}-L|<\epsilon$; in the second case $k \ge K_2$ so $|a_n-L|=|a_{2k+1}-L|<\epsilon$.
Part (b) is fine.
