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Q1: If $\{a_n^2\}$ is convergent, $\{a_n\}$ is convergent

I'm asked to determine whether this is true, and then either prove it or provide a counter example.

I'd say that, as a blanket statement, this is false. My first reaction for this problem would be to use the Monotonic Convergence theorem.

Since $a_n^2$ is positive and always greater than $a_n$, it would need to be an upper bound, and the theorem would only be satisfied if $\{a_n\}$ is increasing, but we're given no information on whether $\{a_n\}$ is increasing so the statement would have to be false.

The next question is:

Q2: If $\{a_n^2\}$ is convergent, and $a_n > 0$, then $\{a_n\}$ is convergent

I would say that this is also false. Just because $a_n > 0$, we don't know that it's increasing still, right?

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    A counterexample could be $a_n=(-1)^n$ – 2017-02-14
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    The first one is indeed false, but not for the reason you give (and observe that $a_n^2 < a_n$ if $0 < a_n < 1$. Can you give a counterexample? (hint: the easiest one alternates). The second one is indeed true, use the fact that $n \mapsto \sqrt{n}$ is a continuous function on $[0, \infty)$. – 2017-02-14
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    The Monotonic Convergence Theorem does not prove that a sequence **isn't** convergent. To prove the claim false, you need to ensure that (in some case) $a_n$ is actually divergent. It makes no sense to justify this by saying "I don't know if I can apply this theorem, therefore it's divergent." – 2017-02-14

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You claim that $a_{n}^{2}>a_{n}$ which is not true.

As to your question, consider $a_{n}=(-1)^n$. Then $a_{n}^{2}=1$ for any $n$. Is this a counter-example?

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It is wrong. Take $(a_n)_n$, where $$a_n=(-1)^n \qquad (n\in \mathbf N).$$