To find $c$ such that $f$ is continuous at $x=0$, we need to find $c$ such that $$\lim _{x\to 0} f(x) = f\Bigl (\answer [given]{0}\Bigr ).$$ In this case, in order to compute the limit, we will have to compute two one-sided limits, since the expression for $f(x)$ if $x<0$ is different from the expression for $f(x)$ if $x>0$. So, $$\begin{align*} \lim_{x\to 0^-} f(x) &= \lim_{x\to 0^-}\answer[given]{\frac{x}{x-1}}\\ &= \frac{\answer[given]{0}}{-1}\\ &=\answer[given]{0}. \end{align*}$$ and $$\begin{align*} \lim_{x\to 0^+} f(x) &= \lim_{x\to 0^+}\left(e^{-x}+c\right)\\ &= e^{\answer[given]{0}} + c\\ &= \answer[given]{1+ c} \end{align*}$$ In order for $f$ to be continuous, the limit $\lim _{x\to 0} f(x)$ has to exist. This means that $$1 + c = 0\qquad \text {so}\qquad c = \answer [given]{-1}.$$ So, if $c=-1$, the limit $\lim _{x\to 0} f(x)=0$. Now, we have find the value

$$f(0) = \answer [given]{0}.$$ Therefore, $$\lim _{x\to 0} f(x) = f\Bigl (0\Bigr )\checkmark ,$$ which proves that $f$ is continuous at $x=0$.

This problem is more challenging because we have more unknowns. However, be brave intrepid mathematician. To find $a$ and $b$ that make $f$ continuous at $x=1$, we need to find $a$ and $b$ such that $$\lim _{x\to 1} f(x) =f(\answer [given]{1}).$$ Since $f(1)=a+b+2$, it follows that $$\lim _{x\to 1} f(x) =a+b+2.$$ We have to compute two one-sided limits, since the expression for $f(x)$ if $x<1$ is different from the expression for $f(x)$ if $x>1$. Looking at the limit from the left, we have $$\begin{align*} \lim_{x\to 1^-} f(x) &= \lim_{x\to 1^-} \left(\answer[given]{x+4}\right) \\ &=\answer[given]{5}. \end{align*}$$ Looking at the limit from the right, we have $$\begin{align*} \lim_{x\to 1^+} f(x) &= \lim_{x\to 1^+} \left(ax^2+bx+2\right) \\ &= \answer[given]{a+b+2}. \end{align*}$$ Hence, for the limit $\lim _{x\to 1} f(x)$ to exist, we must have that $$\begin{align*} 5 &= a+b+2\\ or\\ a+b &= \answer[given]{3}. \end{align*}$$ Also, Hmmmm. More work needs to be done.

To find $a$ and $b$ that make $f$ is continuous at $x=3$, we need to find $a$ and $b$ such that $$\lim _{x\to 3} f(x) =f(3).$$ Since $f(3)=18+a-b$, it follows that $$\lim _{x\to 3} f(x) =18+a-b.$$ Looking at the limit from the left, we have

$$\begin{align*} \lim_{x\to 3^-} f(x) &= \lim_{x\to 3^-} \left(ax^2+bx+2\right) \\ &=a\cdot 9 + b\cdot 3 + 2. \end{align*}$$ Looking at the limit from the right, we have $$\begin{align*} \lim_{x\to 3^+} f(x) &= \lim_{x\to 3^+} \left(6x+a-b\right) \\ &= \answer[given]{18+a-b}. \end{align*}$$ Hence $$\begin{align*} 9a + 3b + 2 &= 18+a-b\\ 8a + 4b -16 &= 0\\ 2a + b -4 &= 0 \end{align*}$$ So now we have two equations and two unknowns: $$a+b=3 \qquad \text {and}\qquad 2a + b = 4.$$ Set $b = 3-a$ and write $$\begin{align*} 2a+3-a&=4 \\ a &= 1, \end{align*}$$ hence $$a=\answer [given]{1}\qquad \text {and so}\qquad b =\answer [given]{2}.$$ Let’s check, so now plugging in values for both $a$ and $b$ we find $$f(x) = \begin {cases} x+4 &\text {if $x<1$,}\\ \answer [given]{x^2+2x+2} &\text {if $1\le x< 3$,}\\ \answer [given]{6x-1} &\text {if $x\ge 3$.} \end {cases}$$ Now $$\lim _{x\to 1^-} f(x) =\lim _{x\to 1^+}f(x) =f(1) = 5,$$ and $$\lim _{x\to 3^-} f(x) =\lim _{x\to 3^+}f(x) =f(3) = 17.$$ So setting $a= \answer [given]{1}$ and $b=\answer [given]{2}$ makes $f$ continuous at $x=1$ and $x=3$. We can confirm our results by looking at the graph of $y=f(x)$: $$\graph [xmin=-10,xmax=10,ymin=0,ymax=30]{y=x+4\left \{x<1\right \},y=x^2+2x+2\left \{1\leq x<3\right \},y=6x-1\left \{3 \leq x\right \}}$$