A few observations to simplify things:
The questions are in terms of \$\frac{dU_A}{dU_E}\$, which is the slope of the transfer characteristic, representing gain only, and ignoring any offset (y-intercept on a graph of \$U_E\$ vs \$U_A\$).
The amplifier is non-inverting, since the input manipulates the non-inverting input.
The amplifier operates in two regimes, X) where the diode is perfectly conducting, Y) where the diode is reverse biased and effectively absent.
When the diode is reverse biased (regime Y), its anode potential is less than the cathode. We can see that further decreases in input potential can only decrease diode anode potential (due to the resistor divider R1, R2) and the diode will remain reverse biased. This also means that the cathode potential is fixed and irrelevant to calculations.
When the diode is forward biased (regime X), its cathode potential becomes dynamic, and can no longer be neglected.
To find the input potential which will just place the voltage across D1 at 0V, at the threshold of transition from regime X to regime Y, it is sufficient to remove D1, and find the input potential which would produce 0V across the bridge between the two dividers R1,R2 and R3,R4. Calling the potential at the junction of R1,R2 \$U_P\$, and at R3,R4 \$U_Q\$:
$$
\begin{aligned}
U_Q &= U_B \frac{R_3}{R_3+R_4} \\ \\
&= U_B \frac{200k\Omega}{200k\Omega+800k\Omega} \\ \\
&= \frac{U_B}{5} \\ \\
\\
U_P &= U_A \frac{R_1}{R_1+R_2} \\ \\
&= U_A \frac{400k\Omega}{400k\Omega+800k\Omega} \\ \\
&= \frac{U_A}{3}
\end{aligned}
$$
\$U_P\$ and \$U_Q\$ are equal at the cross-over point between regimes:
$$ U_P = U_Q = \frac{U_B}{5} $$
To find the input voltage \$U_E\$ that corresponds to this point, which I'll call \$U_{EX}\$ you have one dead simple answer, and two less trivial solutions, depending on how observant you are. Easiest first:
Use the fact that \$u_D = 0V\$, due to opamp action with negative feedback:
$$
\begin{aligned}
U_{EX} &= U_P = U_Q\\ \\
&= \frac{U_B}{5}
\end{aligned}
$$
Use the fact that gain is the reciprocal of feedback factor:
You might have noticed that the feedback factor β is \$\frac{1}{3}\$, from the relationship we derived earlier:
$$ U_P = \frac{U_A}{3} $$
Gain is therefore 3, the reciprocal of \$\frac{1}{3}\$. Using this gain, we can say that:
$$
\begin{aligned}
\frac{U_A}{U_E} &= 3 \\ \\
U_{EX} &= \frac{U_{A}}{3} \\ \\
&= \frac{3U_P}{3} \\ \\
&= U_P \\ \\
&= U_Q \\ \\
&= \frac{U_B}{5}
\end{aligned}
$$
Use the gain equation:
If neither of the above shortcuts work for you, you can just fall back on the good old expression for gain, and follow the same procedure to derive \$U_{EX}\$:
$$
\begin{aligned}
\frac{U_A}{U_E} &= 1 + \frac{R_2}{R_1} \\ \\
&= 1 + \frac{800k\Omega}{400k\Omega} \\ \\
&= 3 \\ \\
\end{aligned}
$$
So, the input voltage at the point of transition from one regime to the other is \$U_{EX} = \frac{U_B}{5}\$. For completeness, the output voltage at that same point, \$U_{AX}\$, is:
$$
\begin{aligned}
U_{AX} &= 3U_{EX} \\ \\
&= \frac{3U_B}{5}
\end{aligned}
$$
For \$U_E \le \frac{U_B}{5}\$ (regime Y) the diode is reverse biased, and the gain equation is simply:
$$ \frac{dU_A}{dU_E} = 3 $$
That was the easy part. Now onto regime X where we may not ignore D1, R3 and R4. There are a few ways of attacking the problem, such as superposition (ughh), nodal/mesh analysis (meh), and reduction by Thevenin's theorem.
To apply Thevenin's theorem, we'll assume the diode is ideal, forward biased, and therefore is effectively a wire link across the bridge between the two dividers. The resistor network starts and ends like this:
simulate this circuit – Schematic created using CircuitLab
Let me redraw the whole circuit again, this time with the Thevenin equivalent derived above:
simulate this circuit
Because of observation #1 at the beginning of this answer, I can confidently ignore \$V_{TH2}\$. All it does is introduce an offset into the gain equation, but I am only after the gain \$\frac{dU_A}{dU_E}\$. So, for \$U_E > \frac{U_B}{5}\$ that will be:
$$
\begin{aligned}
\frac{dU_A}{dU_E} &= 1 + \frac{R_2}{R_{TH2}} \\ \\
&= 1 + \frac{800k\Omega}{\frac{800}{7}k\Omega} \\ \\
&= 8
\end{aligned}
$$
If it's not clear why you can ignore \$V_{TH2}\$, then just figure out the transfer characterstic of the above simplified amplifier:
$$
\begin{aligned}
U_A &= (U_E - V_{TH2}) (1 + \frac{800k\Omega}{\frac{800}{7}k\Omega}) \\ \\
&= U_E\cdot \overbrace{(1 + \frac{800k\Omega}{\frac{800}{7}k\Omega})}^{\text{gain}} \space \overbrace{ - V_{TH2} \cdot (1 + \frac{800k\Omega}{\frac{800}{7}k\Omega})}^{\text{constant offset}}
\end{aligned}
$$
The term \$- V_{TH2} \cdot (1 + \frac{800k\Omega}{\frac{800}{7}k\Omega})\$ is constant. If you take the derivative of \$U_A\$ with respect to \$U_E\$, that term disappears, leaving you with only the coefficient of \$U_E\$, which is gain.
That's my approach to this problem, which I hope was useful. I'll leave you to plug in \$U_B = 10V\$, and see if my answers agree with yours.
To finish with, here's a CircuitLab representation, and the output curve I get when simulating:
simulate this circuit
\$to start and end inline Mathjax to avoid confusion when talking about dollar amounts. (We still use$$for display equations) \$\endgroup\$