HINT: Write $n = p_1^{a_1} \dots p_k^{a_k}$, where $p_1, \dots, p_k$ are distinct primes. Then the number of positive divisors of $n$ is given by $(a_1 +1)(a_2 + 1)\dots(a_k+1)$. In order for this number to be odd, each of the terms $a_i + 1$ must also be odd. What does this tell us about each $a_i$?
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.
Best Answer
I think there is one assumption that you are implicitly using: The prime factorization of $n$ could have more than one prime - let us say $p$ and $q$ - with an odd exponent. In this case you'd have to prove for instance that $p^{1/2} q^{1/2} \not \in \mathbb Z$.