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 \(\boxed{1 \text{ prime number}}\)