Interval(s) where speed is increasing

Differentiate to get $v(t) = s'(t) = t^2 - 4t + 3 = (t-1)(t-3)$ and $a(t) = v'(t) = 2t - 4$. The velocity is zero at $t=1$ and $t=3$, and the acceleration is zero at $t=2$. These three critical values split $[0,4]$ into the subintervals $[0,1]$, $[1,2]$, $[2,3]$, and $[3,4]$.

Check signs on each subinterval: on $[0,1]$, $v>0$ and $a<0$ (opposite signs, slowing down); on $[1,2]$, $v<0$ and $a<0$ (same sign, speeding up); on $[2,3]$, $v<0$ and $a>0$ (opposite signs, slowing down); on $[3,4]$, $v>0$ and $a>0$ (same sign, speeding up).

Therefore the speed is increasing precisely when $v$ and $a$ share the same sign, which happens on $(1,2)$ and $(3,4)$.