Notice that $\rm P(x)=\prod_{k=1}^5(x-r_k)$ and $\rm Q(x)=x^2+1=(x-i)(x+i)$. So $\rm \prod_{k=1}^5Q(r_k)= \prod_{i=1}^5 [(r_k-i)(r_k+i)]= \prod_{k=1}^5 (r_k-i) \prod_{k=1}^5(r_k+i)$ $\rm =[-P(i)][-P(-i)]=P(i)P(-i)=(i^5-i^2+1)((-i)^5-(-i)^2+1)=(i+2)(2-i)=2^2-i^2=5$

Assume the roots of the polynomial $P(x)=x^5-x^2+1$ are $\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5.$ We have to find the value of $Q(\alpha_1)Q(\alpha_2)Q(\alpha_3)Q(\alpha_4)Q(\alpha_5)$ where $Q(x)=x^2+1.$ Therefore we get, $$\prod_{i=1}^{5}Q(\alpha_i)=\prod_{i=1}^{5}(\alpha_i^2+1)\implies \prod_{i=1}^{5}(\alpha_i+i)(\alpha_i-i)$$Now plugging $x=i$ and $x=-i$ in $P(x)$ and multiply it we get, $$P(i)\cdot P(-i)=(2+i)(2-i)\implies\boxed{5.}$$