Answer
$(a)$ $15^{th}$ St. and $12^{th}$ Ave.
$(b)$ Each of them has to walk $17$ $blocks$
Work Step by Step
For a better visualization, let's put these points on a coordinate plane (Refer to the image above).
Let $A$ be the house of first friend, located at the corner of $3^{rd}$ St. and $7^{th}$ Ave. So we will get point $A(3, 7)$
Using the same idea, the house $B$ of the other friend located at the corner of $27^{th}$ St. and $17^{th}$ Ave. will be $B(27, 17)$.
The coffee shop $C$ is located on the middle of the distance between their houses. Using mid-point formula $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$, we can locate the coffee shop.
$(\frac{3+27}{2}, \frac{7+17}{2})$ =>$(\frac{30}{2}, \frac{24}{2})$ => $(15, 12)$
That is the corner of $15^{th}$ St. and $12^{th}$ Ave.
$(b)$ We have to calculate walking distance for each of them. Using distance formula: $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
The friend from $A$ has to walk $(AD+DC)$, while the friend from $B$ has to walk $(BE+EC)$.
$AD=\sqrt{(15-3)^2+(7-7)^2}=\sqrt{12^2}=12$
$DC=\sqrt{(15-15)^2+(12-7)^2}=\sqrt{5^2}=5$
Friend from $A$ has to walk $12+5=17$ $blocks$
$BE=\sqrt{(27-27)^2+(12-17)^2}=\sqrt{(-5)^2}=\sqrt{25} = 5$
$EC=\sqrt{(15-27)^2+(12-12)^2}=\sqrt{(-12)^2}=\sqrt{144} = 12$
Friend from $B$ has to walk $5+12=17$ $blocks$
