Perfect square product among 17 integers
For a)
Let's $M$ be the set of that $17$ numbers and make a map $\phi : M\to \mathbb{Z}_2^4$ such that each $$2^a3^b5^c7^d \mapsto (a_{\pmod 2}, b_{\pmod 2}, c_{\pmod 2}, d_{\pmod 2})$$ Then since $|\mathbb{Z}_2^4|=16$ we have $m\ne n$ such that $\phi (m) =\phi (n)$, thus $m\cdot n $ is a perfect square.
So we assign to each number $2^a3^b5^c7^d$ in $M$ a $4$-tuple $(a',b',c',d')$ where $x'$ is $0$ if $x$ is even and $1$ if $x$ is odd, for each $x\in\{a,b,c,d\}$. Now since the number of all such $4$-tuples is $16$ and we have $17$ numbers, two of them must have the same $4$-tuple and therefore their product is a perfect square.
Solution for b): For a solution we will use a). Say a pair of numbers $x,y$ is good if their product is a perfect square.
From $49$ numbers take any $17$ numbers, then we have a good pair $a_1,a_2$ among them.
Now from the rest of the $47$ numbers take any $17$ numbers, then we have a good pair $a_3,a_4$ among them.
Now from the rest of the $45$ numbers take any $17$ numbers, then we have a good pair $a_5,a_6$ among them.
and so on. We repeat this process until we get last good pair $a_{33},a_{34}$ (and we are left wit $15$ numbers so that we can't repeat the process).
Now calculate their products $b_i:=a_{2i-1}a_{2i}$. So we have $17$ perfect squares $b_1,b_2,...b_{17}$. Each $b_i$ we can write like this
$$b_i = 4^x 9^y 25^z 49^t$$
Again we assign to each $b_i$ $4$-tuple $(x',y',z',t')$ where $w'$ is remainder of $w$ modulo $2$. Again two of the $b_i$ must have the same $4$-tuple, since we have $17$ numbers and only $16$ $4$-tuples. So, their product is perfect $4$-power.
This approach is not different from the others proposed, but I wanted to illustrate a point which may help you to express yourself.
The point is that we can start with a definition, and if we are clear about that, other things fall more easily into place.
So if $n=2^a3^b5^c7^d$ we define the signature of $n$ to be $(a, b, c, d) \bmod 2$ meaning that each entry in the signature is $0$ or $1$ depending on whether each of $a, b, c, d$ is respectively even or odd.
There are sixteen $(2^4)$ possible signatures, so amongst the seventeen numbers two of the signatures must be equal. When you multiply two numbers with equal signatures together you get a number with zero signature. And a number with zero signature has every exponent even, and therefore must be a square.
I wanted to illustrate what was in my comment, and also to show that if there is something slightly complicated, giving it a name means you only have to deal with the complications once.
For the second bit, John Watson has shown how to take out seventeen pairs which multiply to give a square. You might use the factorisation $4^a9^b25^c49^d$ to define the "second signature" or some such, and the same argument essentially goes through to give you a fourth power.