EnormousBighead

In a right triangle the median has the same length as half the hypotenuse because of the circumcircle of the triangle. Therefore:

$\triangle{ZCA}$ is isoceles and $\angle{ZAC}=\angle{ZCA}=90-\angle{ABX}$

$\angle{ABX}=90-\angle{BAX}=90-(\frac{\angle{BAC}}{2}-\angle{XAY})=90-(45-22)=67$

Thus $\angle{ZAC}=\angle{ZCA}=90-67=\boxed{23}$

$\frac{7!}{3!\times2!\times2!}=\boxed{210}$

Let the numbers be $x-3,x-2,x-1,x,x+1,x+2,x+3$
Their sum is
 $7x=\frac{4}{3}(x+3)$

Solving we get:
$x=\frac{12}{17}$ 
The largest number is $x+3=\frac{12}{17}+\frac{3}{1}=\boxed{\frac{63}{17}}$
Which is not an integer and the question is flawed. 

$Q$ lies on the line $y=3x$
$P$ lies on the line $y=x$

WLOG lets $Q=(1,3)$, therefore $OP=OQ=\sqrt{10}$

Thus we can say $P=(\sqrt{5},\sqrt5)$

Slope between 2 points is $s=\large{\frac{y_2-y_2}{x_2-x_1}=\frac{\sqrt5-3}{\sqrt5-1}=\boxed{\frac{1-\sqrt5}{2}}}$

A number with $3$ positive integers must be in the form $p^2$. Therefore the answer is only $\boxed{1 \text{ prime number}}$