Answer
See Graph (Ordered Pairs)
Yes - positively linear
$y=0.55x+195.078$
$\approx 786$
Work Step by Step
While Part A is the plotted points on the graph with half of Part B on the graph as well as the linear regression line, which is also known as the "line of best fit", Part A already implied for Part B that the data is appearing to be positively linear.
So, one method for finding the linear regression is using the closest middle two points and find the equation that way. If that is the case, then I chose the 3rd and 3rd to last ordered pairs, which are $(1043,770)$ and $(1248,883)$. You can find the slope by doing the following:
$$m=\frac{y_2-y_1}{x_2-x_1}=\frac{883-770}{1248-1043}=\frac{113}{205}=0.551219512$$
Now for the function iself, which can be solved by the following:
$$\begin{matrix}
y=mx+b\\
y=\frac{113}{205}x+b\\\\
770=\frac{113}{205}(1043)+b\\\\
770=\frac{117859}{205}+b\\\\
770=\frac{117859}{205}+b\\\\
770-\frac{117859}{205}=\frac{117859}{205}+b-\frac{117859}{205}\\\\
\frac{770\cdot 205}{205}-\frac{117859}{205}=b\\\\
\frac{157850}{205}-\frac{117859}{205}=b\\\\
\frac{157850-117859}{205}=b\\\\
\frac{39991}{205}=b\\\\\\
y=0.551219512x+195.078049\\
y=0.55x+195.078
\end{matrix}$$
Part C is basically just plug in play, while estimating to a whole number. So, with $x=1075$, you can say the following:
$$\begin{matrix}
x=1075\\
y=y(x)=0.55x+195.078\\
y(1075)=0.55(1075)+195.078\\
y(1075)=591.25+195.078\\
y(1075)=786.328\\
y(1075) \approx 786
\end{matrix}$$
