1. We are given $F(x)=\int_0^x t f(t)\,dt$ and $F(x^2)=x^4+x^5.$

We need to find $\sum_{r=1}^{12} f(r^2).$

2. Let $u=x^2.$ Then the given relation becomes $F(u)=u^2+u^{5/2} \qquad (u>0),$ because $x^4=(x^2)^2=u^2, \qquad x^5=(x^2)^{5/2}=u^{5/2}.$ So, $F(x)=x^2+x^{5/2}.$

3. Differentiate $F(x)$ using the Fundamental Theorem of Calculus: $F'(x)=x f(x).$ Also, from $F(x)=x^2+x^{5/2},$ we get $F'(x)=2x+\frac{5}{2}x^{3/2}.$ Hence, $x f(x)=2x+\frac{5}{2}x^{3/2}.$ For $x>0$, divide by $x$: $f(x)=2+\frac{5}{2}x^{1/2}.$

4. Therefore, $f(r^2)=2+\frac{5}{2}\sqrt{r^2}=2+\frac{5}{2}r$ since $r\ge 1$.

5. Now compute the sum: $\sum_{r=1}^{12} f(r^2)=\sum_{r=1}^{12}\left(2+\frac{5}{2}r\right).$ Break it into two sums: $=\sum_{r=1}^{12}2+\frac{5}{2}\sum_{r=1}^{12}r$ $=24+\frac{5}{2}\cdot \frac{12\cdot 13}{2}.$ Now, $\frac{12\cdot 13}{2}=78,$ so $24+\frac{5}{2}\cdot 78=24+195=219.$

6. Final answer: $\sum_{r=1}^{12} f(r^2)=219.$