Linear Regression about SSR I've studied about Anova.
But I wonder 
$$SSR=\sum_{i=1}^{30}(\hat{y}_i-\bar{y})^2=(b_1)^2\sum_{i=1}^{30}(x_i-\bar{x})^2$$
In this equation, why does $(b_1)^2$ appear?? and how to tranform summation about y to it about x?
And finally Why do they use $b_1$ instead of $β_1$
 A: Are you sure that this is the sum of squares of residuals?  I would have expected $\displaystyle \sum_{i=1}^{30}(y_i - \hat{y}_i)^2$
Meanwhile you have something like $\hat{y}_i= b_0 + b_1 x_i$ and  $\bar{y}=b_0 + b_1 \bar{x}$ from your linear regression, so $\hat{y}_i-\bar{y}=b_1(x_i-\bar{x})$, giving $$\displaystyle \sum_{i=1}^{30}(\hat{y}_i-\bar{y})^2 = \sum_{i=1}^{30}(b_1(x_i-\bar{x}))^2 = b_1^2 \sum_{i=1}^{30}(x_i-\bar{x})^2$$ but this is what I would expect ANOVA to call the sum of squares of treatment
A: As I was typing, Henry has provided the answer and pointed out a possible typo. Anyway here it goes:
Usually the use of $b_0$ (for the intercept) and $b_1$ (for the slope) is meant to be less intimidating, in an intro level material. This is exactly the same as using $\beta_i$, but I personally almost never see the use of $b_i$ for $i \geq 2$. 
Here presumably the linear model is $\hat{y}_i = b_0 + b_1 x_i$ with the equation saying "the line shall pass through the center of the data":
$$\overline{y}=b_0 + b_1 \overline{x} \tag*{Eq.(1)}$$
You'll get the right hand side by substituting this into SSR and rearranging it (using the fact that $\sum_i{(x_i - \overline{x} )= 0}$  at some point).
My opinion is that you don't need to "memorize" this for the slope 
$$b_1 = \frac {\sum _{i=1}^{n}(x_{i}-{\bar {x}})(y_{i}-{\bar {y}})}{\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}}$$
which is the other of the two equations (for solving the 2 unknowns $b_0$ and $b_1$), but yeah you should keep in mind of the above Eq.(1) about the line passing through the center.
