Geometrical proof required regarding midpoints $ABCD$ is a convex quadrilateral with $W, X, Y, Z, M$ and $N$ as the midpoints of $AB, BC, CD, DA,$ the diagonals $AC$ and $BD$ respectively. [Then, $WXYZ$ is a parallelogram with $K$ as the intersection point of its diagonals by Varignon's Theorem]



The 1st question is: “Will $K$ be the center of the circle using $MN$ as diameter?”

Line $BP$ is now drawn parallel to $AC$ cutting $DC$ produced at $P$. Let $Q$ be the midpoint of $DP$. $MN$ is produced to cut $BC$ at $R$. 

The 2nd question is: “Is $RQ \parallel  AD$?”

PS: The red and green lines are not perpendicular to each other.

Picture added to reflect @Aretino 's answer to the first question.
To the first question: yes, because:
$$
K={1\over2}(W+Y)=
{1\over2}\left({A+B\over2}+{C+D\over2}\right)=
{1\over2}\left({A+C\over2}+{B+D\over2}\right)=
{1\over2}(M+N).
$$
More geometrically.

In triangle $CAD$, $M$ and $Y$ are midpoints of two sides, so we have $MY=AD/2$ and $MY\parallel AD$. In the same way, in triangle $ABD$ we have $NW=AD/2$ and $NW\parallel AD$. It follows that $MY=NW$ and $MY\parallel NW$, but then $MYNW$ is a parallelogram and its diagonals $MN$ and $WY$ have the same midpoint $K$.
Regarding the second question, note that the answer is no, in general, due to the following remark:

Two vectors $u, v$ are such that $u, v$ have the same length if and only if the vectors $u+v, u-v$ are orthogonal.

The result follows from calculating $(u+v)\bullet (u-v)$ right away.

Taking $u=\overrightarrow{KW}, v=\overrightarrow{KZ}$ one has: $u+v=\overrightarrow{XW}$, and $u-v=\overrightarrow{YX}$, so the only possibility that your parallelogram is inscribed in a circle centred at $K$ is for it to be a rectangle.

I'm sure that you can draw a rhombus or some other parallelogram, and then find a convex quadrilateral so that it all fits your picture, then you'll have the desired counterexample.  In fact, it suffices to take the above example in the figure and to apply an affine transformation which does not preserve angles!

The proof that your figure is a parallelogram uses simply baricentric coordinates, or vectors, so it is an affine concept which is therefore preserved by affine transformations, which in coordinates have the form

$$x'=ax+by+e, y'=cx+dy+f,$$

but the condition that the figure be a rectangle is easily broken by a convenient affine transformation, if we place the axes so that none of them is parallel to the sides of your rectangle and then take, say $x'=3x, y'=y.$ 

Addition: It is known that for every two bases of $\mathbb{R}$, $u, v$ and $u', v'$, there is a $2\times 2$-matrix $A$ such that $Au=u', Av=v'.$

This clearly means that we can send two perpendicular vectors to a non-perpendicular pair holding an arbitrary angle between them.  Hence, the affine-geometric construction shown will certainly have an arbitrary angle between vectors $u, v$ defined as above.

Hope this is clear enough.