To discuss differentiability of function at origin and my attempt P1:  $F =  |x| +  |y|  when x,y is not equal to 0,

  =   0   when x = y = 0    


P2 :Discuss the differentiability at origin  of  $F = y sin(1/x)$
 :
Near good, but for $|x|+|y|$. Do not use argument, that partial derivatives are not continuous. Theorem says "if partial derivatives... are continuous... derivative exists", but not reverse. However argument with one sided partial derivatives in enough.

For the corrected version of P2 ($f(x,y)=y\sin(1/x)$ for $x\neq0$ and 0 for $x=0$):
$$
f'_y(0,0)=\lim_{y\to0}\frac{0-0}{y}=0,\quad
f'_x(0,0)=\lim_{x\to0}\frac{0\sin(1/x)-0}{x}=0,
$$
hence (in case P2) partial derivatives exist.

Function $f$ is differentiable in $(x_0,y_0)$ if
$$
\lim_{(h,k)\to(0,0)}\frac{f(x_0+h,y_0+k)-f(x_0,y_0)-f'_x(x_0,y_0)h-f'_y(x_0,y_0)k}{\sqrt{h^2+k^2}}=0.
$$
In our case we obtain
$$
\lim_{(h,k)\to(0,0)}\frac{k\sin(1/h)-0-0-0}{\sqrt{h^2+k^2}}.
$$
Let $k=h$. Then
$$
\lim_{h\to0}\frac{h\sin(1/h)}{\sqrt{2h^2}},
$$
which doesn't exist.

Edit: I can see, that there is yet another definition of $f$ ($f(0,y) = y$). Then $f'_y(0,0)=1$ and modifications are the following:
$$
\lim_{(h,k)\to(0,0)}\frac{k\sin(1/h)-0-0-k}{\sqrt{h^2+k^2}}.
$$
Let $k=h$. Then
$$
\lim_{h\to0}\frac{h(\sin(1/h)-1)}{\sqrt{2h^2}},
$$
which again doesn't exist.

Phew!
For the first question, your answer is wrong. The limits you've given do exist at the origin, the problem is that they come to different numbers depending on the direction. Thus the derivative is not well-defined.