Is a number meeting these conditions divisible by forty-nine?

You can derive this fairly directly from Euclid's Lemma, which says that if a product $a \cdot b$ is divisible by prime $p$, then either $a$ or $b$ (or both) is divisible by $p$.

So, $n^2$ divisble by 7 means $n \cdot n$ divisible by 7 which (by Euclid's Lemma) means $n$ divisible by 7 and thus $n^2$ divisible by 49.

Of course it took even Euclid a little work to prove the Lemma

Yes.

To see this, call the number $x$ and suppose that $x$ is the square of the number $y$. Suppose that $y$ has a prime factorization

$$y = p_1^{n_1}p_2^{n_2}\cdots p_K^{n_K}$$

where each $n_k>0$ and $p_1<p_2<\cdots < p_K$. (In other words, we're displaying the prime factorization as compactly as possible, and "in order".)

Since $x = y^2$, we have

$$x = (p_1^{n_1}p_2^{n_2}\cdots p_K^{n_K})^2 = p_1^{2n_1}p_2^{2n_2}\cdots p_K^{2n_K}$$ Note all of the powers $2n_1,2n_2,\ldots,2n_K$ are even; and since $7$ divides $x$, one of the prime factors $p_i$ must be $7$.

But then $2n_i$ is an even number greater than zero, so is at least $2$. Thus there are at least two factors of $7$ in $x$, so $49$ divides $x$.

Yes, it is true. It hinges on the fact that $7$ is a prime number.

In general, if $n$ is an integer that is divisible by a prime number $p$ and $n$ is a square, then $n$ is divisible by $p^2$.

This follows from the Fundamental theorem of arithmetic.