Exercise

Correct: The tangent line is the form $ax + b$ and goes through the point $(0,f(0)) = (0, (2+0)e^0) = (0,2)$. We know that $a=f'(0)$. Hence, we have to determine $f'(x)$. Write $f(x) = u(x)\cdot v(x)$ with $u(x) = 2+x$ and $v(x) = e^x$. Then it holds that (see possibly Derivatives of elementary functions):
$$\begin{align*}
u'(x) &= 0 + 1 = 1\\
v'(x) &= e^x\\
f'(x) &= u'(x)v(x) + u(x)v'(x) = 1\cdot e^x + (2+x)e^x = (3+x)e^x\\
a=f'(0) &= (3+0)e^0 = 3.
\end{align*}$$
The value of $b$ can now easily be found, because $(0,2)$ is a point on the tangent line:
$$
\begin{align*}
3\cdot 0 + b &= 2\\
b&= 2.
\end{align*}$$
Hence, the equation of the tangent line is $3x + 2$.

Go on.

Wrong: You have to give the equation of the tangent line, not the derivative.

See Derivatives of elementary functions: Example 2.

Wrong: You need the product rule to determine the derivative of $f(x)$. Note that if $f(x)=u(x)v(x)$, then
$$ f'(x) \neq u'(x)v'(x).$$

See Product rule.

Wrong: You have a sufficient amount of information.

See Derivatives of elementary functions: Example 2.