Hodgin-Huxley model for a single neuron - continuation
Actually there is a way to determine the initial conditions for this problem.

We assume that the gating variables $r$ and $s$ are at a steady state before the membrane potential jump from$u_0$ to $u'$. 

If we look at the first graph(the steady state relation versus membrane potential) we can see that at the membrane potential $u_0$ that $r(t=0)\approx 0$ and $s(t=0)\approx 1$. 

Then we make our jump in voltage from$u_0$ to $u'$ and now our $r(t)$ and $s(t)$ will change towards their new steady values $r'=1$ $s'=0$ with time constants given by the second graph.

Your equations $r(t) =C_1 e^{-t}+1$ and $s(t)=0+C_2 e^{-t/15}$ are correct, but using the information from the graphs we can se that $C_1= -1$ and $C_2= 1$. 

From these equations we can see that $r(3)=1-e^{-3}\approx .9502$ and $s(3)= e^{\frac{-1}{5}}\approx .81$ (note that .81 is close enough to 1 for this particular problem). This means that the channel is mostly open as statement 4 asserts.

Now for statement 7 we have $t= 100$ and while $r(100)= 1-e^{-100}\approx1$, and $s(100)=e^{\frac{-100}{15}}\approx 0$. This means the product $r(100)^2* s(100)\approx 1^2 * 0 \approx 0$ thus the channel is mostly closed. 

Hope this helps.