Part one:

We rewrite the equation as:

\begin{eqnarray*} {X(t)\over K} &=&
\exp\left(\log\left(\frac{X(0)}{K}\right)\exp(-\alpha t)\right) \\
\log\left(\frac{X(t)}{K}\right) &=&
\log\left(\frac{X(0)}{K}\right)\exp(-\alpha t)\\ \frac{
\frac{\mathrm{d}X(t)}{\mathrm{d}t} }{X(t)} &=&
\log\left(\frac{X(0)}{K}\right)(-\alpha)\exp(-\alpha t))\\
{\mathrm{d}X(t)\over \mathrm{d}t} &=&
\alpha\,X(t)\log\left(\frac{K}{X(t)}\right) \end{eqnarray*}

which is our required differential equation.

Part two:

Again we rewrite our equation as:

\begin{eqnarray*} P(t) &=& \frac{a\exp(bt)}{a-1+exp(bt)}
\\ P(t)(a-1)+P(t)\exp\left(bt\right) &=&
a\exp\left(bt\right) \\ \frac{P(t)(a-1)}{a-P(t)} &=&
\exp\left(bt\right)
\\ \log\left(P(t)(a-1)\right)-\log\left(a-P(t)\right)
&=& bt \\ \Biggr(\frac{(a-1) \mathrm{d}P}{(a-1)P(t)} +
\frac{\mathrm{d}P}{a-P(t)}\Biggr) &=& b\mathrm{d}t \\
\Biggr(\frac{1}{P(t)} + \frac{1}{a-P(t)}\Biggr) \mathrm{d}P
&=& b\mathrm{d}t \\ \frac{\mathrm{d}P}{\mathrm{d}t}
&=& \frac{b}{a} P(t)(a-P(t))

\end{eqnarray*}