-- For $a)$ and $d),$ counter-examples are identity matrices like $M= \begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix}$ for $n=2.$
Because, considering a non-zero vector $x$ as $= \begin{bmatrix} a\\b \end{bmatrix}$ where, $a >0, b>0$
Now, $x^TMx = \begin{bmatrix} a &b \end{bmatrix} \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} a \\b \end{bmatrix} = a^2 + b^2 > 0.$ So, such $Ms$ exist and eigen values can be real.
-- For $(c),$ counter-example is : $M= \begin{bmatrix} 2k &2k \\ -2k & 2k \end{bmatrix}$ where $k \geq 1$
Here, also on considering a non-zero vector $x$ as $= \begin{bmatrix} a\\b \end{bmatrix}$ where, $a >0, b>0$
Now, $x^TMx = \begin{bmatrix} a &b \end{bmatrix} \begin{bmatrix} 2k &2k \\-2k & 2k \end{bmatrix} \begin{bmatrix} a \\b \end{bmatrix} = 2k(a^2 + b^2) > 0.$ So, such $Ms$ exist and eigen values can be complex here.
-- Now, for $(b),$ Suppose, $x \in \mathbb{R}^n$ is a non-zero eigen vector associated with matrix $M$.
Then, $Mx = \lambda x$
$\Rightarrow x^TMx = x^T\lambda x$
$\Rightarrow x^TMx = \lambda x^Tx$
Since, $x^Tx > 0 $ and If $\lambda > 0$ then $x^TMx > 0$
So, For positive eigenvalues $\lambda$ and a non-zero eigen-vectors $x$, $x^TMx > 0$ is always true.
Since, determinant is the product of eigen values. So, if all eigen values are positive then determinant will be positive and matrix $M$ will be non-singular and has full rank $n$. So, $(b)$ is true.