the first part is rather trivial: $q(n)=\sum_\lambda \text{disp}(\lambda)$ as $\lambda$ runs over all the partitions of $n$ and $\text{disp}(\lambda)$ counts the number of distinct parts in $\lambda$. clearly this is the same as counting for each $i$, the number of partitions of $n$ that count a distinct part of size $i$.If $i=n$ then of course there is only one such partition. else, the number of partitions that have a part of size $i\ = p(n-i)$, so $q(n)=1+\sum_{i=1}^{n-1} p(n-i)$ and that is what we need. the second part is easily improved by the fact that $np(n)=\sum_{i=1}^n \sigma(i)p(n-i)$ where $\sigma(i)$ is the sum of all positive divisors of $i$.

I just think the second question should be changed a little. The original problem should be $ 1+\sum_{i = 1}^{n}P(i)\leq\sqrt{2n}P(n)$