This is a typical differential equation problem. It leads to what is known as a "logistics model".
Let $N(t)$ denote the number of infected people at time $t$ after the first infection, with $t$ measured in days. The rate at which the disease spreads is the rate of change of $N(t)$, which is the derivative $N'(t)$. We are told that the rate of change is proportional to $N(t)$ (the number of people who already have the disease) times $10000-N(t)$ (the number of people who don't have the disease). That is:
$$N'(t) = kN(t)(10000-N(t))$$
for some constant $k$.
We are also told that $N(0) = 1$ (one person gets sick) and $N(5)=50$. We are asked for the value of $N(10)$.
Note that we cannot have $N(t)=0$ for all $t$, nor can we have $N(t)=10000$ for all $t$.
This differential equation is of a kind called "separable". We can solve it as follows:
$$\begin{align*}
\frac{dN}{dt} &= kN(10000-N)\\
\frac{dN}{N(10000-N)} &= k\,dt\\
\int\frac{dN}{N(10000-N)} &= \int k\,dt\\
\int\frac{dN}{N(10000-N)} &= kt + C.
\end{align*}$$
for some constant $C$. To solve the integral on the left, we can use Partial Fractions:
$$\frac{1}{N(10000-N)} = \frac{\quad\frac{1}{10000}\quad}{N} + \frac{\quad\frac{1}{10000}\quad}{10000-N}$$
so
$$\begin{align*}
\int\frac{dN}{N(10000-N)} &= \frac{1}{10000}\int\left(\frac{1}{N} + \frac{1}{10000-N}\right)\,dN\\
&= \frac{1}{10000}\left(\ln|N| - \ln|10000-N|\right) + D.
\end{align*}$$
Putting it all together, we get
$$\frac{1}{10000}\left(\ln |N| - \ln|10000-N|\right) = kt + E,$$
where $E$ is some constant. Rewriting, we have:
$$\ln\left|\frac{N}{10000-N}\right| = Kt + F,$$
where $K$ and $F$ are constants. Since $N$ is always between $1$ and $10000$, we can drop the absolute values.
Plugging in $t=0$, we know that $N=1$, so we have
$\ln\frac{1}{9999} = F$. Plugging in $N=5$, we know that $N(t)=50$, so we have
$$\ln\frac{50}{9950} = 50K + \ln\frac{1}{9999}.$$
From here, we can plug in $t=10$ and solve for $N$.
A start: Let $y=y(t)$ be the fraction who have heard by time $t$. Then the fraction who have not is $1-y$. The rate of change of $y$, we are told, is proportional to the product $y(1-y)$. Our differential equation is therefore
$$\frac{dy}{dt}=ky(1-y).$$
This is a special case of the logistic equation, which you know how to solve.
It is convenient to let $t=0$ at $8\colon00$. So $y(0)=\frac{80}{1000}$. We are told that $y(4)=\frac{1}{2}$. These two items are enough to tell us everything about the equation, including the constant $k$. Some algebraic manipulation will be needed.
Now that you have the equation for $y(t)$ in terms of $t$, you can find the $t$ such that $y(t)=0.9$. Note that this $t$ is the time elapsed since $8\colon00$ AM. You will need to give the answer in clock terms.
Best Answer
$$\frac{ds}{dt} = k(100000 - s)$$
$$\frac{ds}{(100000 - s)} = k\ dt$$
$$\ln(100000 - s) = kt + C$$
$$100000 - s = Ce^{kt}$$
$$\text{When}\ t = 0, C = 100000 - 10000 = 90 000$$
$$\text{When}\ t = -7, k = \frac{\ln\frac{10}{9}}{-7} = -.01505$$
$$s = 100000 - 90000e^{-.01505t}$$
$$50000 = 90000e^{-.01505t}$$
$$t = \frac{\ln\frac{5}{9}}{-.01505}$$
$$t = 39\ \text{days}$$