For the first function observe that it is a composition of $(x-1)^2$ and $e^x$ and since both are continuous and moreover, they are differentiable on $\Bbb R$, the composition i.e. $e^{(x-1)^2}$ is differentiable over $\Bbb R$.
So differentiating, $f^/(x) = 2(x-1)e^{(x-1)^2}$ and $$ f^{(2)}(x) = {4(x-1)^2}e^{(x-1)^2} + 2e^{(x-1)^2}$$.
Now equating, $f^/(x) = 0 $ we obtain that x = 1.
Putting x = 1 at $ f^{(2)}(x) = {4(x-1)^2}e^{(x-1)^2} + 2e^{(x-1)^2}$ we deduce that $f^{(2)}(x) > 0$, so f has a global minima at x = 1 . At that point , f(1) = 1.
So at all other points $f(x)>1$ and since $f$ is continous on a connected interval, it satisfies the IVP and thus answer to your first question becomes $[1,+\infty)$ .
[Actually, you can consider the graph of $e^{(x-1)^2}$ to be that of $e^{x^2}$ with the origin right-shifted at x = 1.]
For the second question observe that $2x^2 + 5x - 5$ is a polynomial and so it is also continuous and also since the set you're considering the pre-image of is a connected set you just need to solve the quadratic for the boundary points of the inetrval i.e. you just need to solve
$$ 2x^2 + 5x - 5 = 0$$ and $$2x^2 + 5x - 5 = 3$$
The roots you'll obtain probably, $$x = \frac{-5 \pm \sqrt{65}}{4}$$ for the first one and $$x = \frac{-5 \pm \sqrt{89}}{4}$$ for the second one. Now take suitable union and you'll obtain answer .
[$\frac{-5 + \sqrt{89}}{4}$ will replace 28 in your answer and $\frac{-5 - \sqrt{89}}{4}$ will replace -28]
The method is just solving equations and drawing the lines/planes. Let's see how.
Let's first look at the (pre)image of $f(z)=\exp(z)$ and $g(z)=\log(z)$ in $A=\{z\in \mathbb{C} \mid \mathrm{Im} z=0\}$. We write $z=x+iy$ and note that $A=\{x\mid x\in\mathbb{R} \}=\mathbb{R}$ is just the real line, because $y=\mathrm{Im} z=0$.
Now we easily see that $f(A)=\{f(z)\mid z\in A\}= \mathbb{R}_{>0}$, so $f$ maps $A$ (the real line) to half a real line.
The arrow represent the function sending elements from the arrow-tail set to the arrow-head set.
Now for the preimage $f^{-1}(A)=\{z\mid f(z)\in A\}$, we use that $f(x+iy)=e^xe^{iy}=e^x(\cos(y)+i\sin(y))=e^x\cos(y)+ie^x\sin(y)$ and thus we want $ie^x\sin(y)=0$, which happens for $y=k\pi$ with $k\in\mathbb{Z}$, so the preimage are the lines $x+ ik\pi$, so $f^{-1}(A)=\{x+ ik\pi\mid x\in\mathbb{R},k\in\mathbb{Z}\}$.
Now using that $g(z)=\log(z)=\ln(|z|)+i \text{arg}(z)$ with the argument in $\mathbb{R}$ (this is actually all logarithmic functions $\log(z,t)$ with their argument in $(t,t+2\pi]$ with $t\in\mathbb{R}$, since you didn't specify which argument we work in), we find (note that $0$ is not in the domain of $g$) that $g(A\backslash\{0\})=\{x+ ik\pi\mid x\in\mathbb{R},k\in\mathbb{Z}\}$, because the argument of the real line is $k\pi$ with $k\in\mathbb{Z}$.
If you meant $g(z)=\text{Log}(z)$ with principal argument in $(-\pi,\pi]$, then only $k=0$ and $k=1$ hold:
For any individual logarithmic function $\log(z,t)$, there will only be two neighbouring lines on the right, namely the ones with $y\in (t,t+2\pi]$.
The same way, we find that $g^{-1}(A)= \mathbb{R}_{>0}$, because we want $\mathrm{Im} g(z)=\text{arg}(z)=0$.
Now try the same for $B$: use that $x=\mathrm{Re}z=\mathrm{Im}z=y$, and remember that $$f(x+iy)=e^x(\cos(y)+i\sin(y))$$ and $$g(z)=\text{Log}(z)=\ln(|z|)+i \text{Arg}(z)$$ (I think it's better to use the principal argument). So $f(B)=\{e^x(\cos(x)+i\sin(x))\mid x\in\mathbb{R}\}$, in which we recognize a spiral. $f^{-1}(B)=\{x+i(\frac{\sqrt{2}}{2}+k)\pi\mid k\in\mathbb{Z}\}$, which is a collection of horizontal lines. This is because we need to find solve $e^x\cos(y)=\mathrm{Re}f(z)=\mathrm{Im}f(z)=e^x\sin(y)$, so $y=\frac{\sqrt{2}}{2}+k\pi$.
You do not have to do every part. I just need some assistance.
I hope that it's clear now how to do this. Good luck with the other ones.
Best Answer
I also agree with your instructor.
$1)$ The graph of $f$ is a parabola with minimal point at $f(2)=-5$ and up concavity. See that $f(1)=-4$:
$f(x)=1 \Rightarrow x^2-4x-1=1 \Rightarrow x=2+\sqrt{6}$ and $x=2-\sqrt{6}$
That means if we take $x<2-\sqrt{6}$ or $x>2+\sqrt{6}$ we will get $f(x) \in (1,\infty)$.
$2)$ If $|z|<1$ write $z=r(\cos\alpha+i\sin\alpha)$, with $r<1$ and $\alpha \in [0,2\pi]$ and then $g(z)=z^5=r^5(\cos5\alpha+i\sin5\alpha))$ with, $r^5<1$ and $5\alpha \in [0,5\pi]$.