I doubt that this is the best approach since it basically only uses the definition of positive semidefinite (PSD) matrices and requires a fact about the pointwise product of PSD matrices, but maybe it will give you some ideas.
We have that,
$$
\sin(x_i-x_j)^2 = \frac{1}{2} - \frac{1}{2}\cos(2(x_i-x_i))
= \frac{1}{2} - \frac{1}{2}\big[ \cos(2x_i)\cos(2x_j) + \sin(2x_i)\sin(2x_j) \big]
$$
Therefore we can write,
$$
A_{i,j} = e^{-\lambda/2}\exp\left(\frac{\lambda}{2}\cos(2x_i)\cos(2x_j) +\frac{\lambda}{2}\sin(2x_i)\sin(2x_j) \right)
$$
Now, note that the matrix,
$$
B_{i,j} = \frac{\lambda}{2}\cos(2x_i)\cos(2x_j) + \frac{\lambda}{2}\sin(2x_i)\sin(2x_j)
$$
is the sum of two rank-1 outer products and therefore PSD.
Note: Maybe for your needs its sufficient to show the part of the kernel in the exponent is the sum of separable functions, in which case you are done.
Using Taylor expansion we have,
$$
A_{i,j}
%= e^{-\lambda/2} \sum_{k=0}^{\infty} \frac{1}{k!} \left( \frac{\lambda}{2}\cos(2x_i)\cos(2x_j) + \frac{\lambda}{2}\sin(2x_i)\sin(2x_j) \right)^k
= e^{-\lambda/2} \sum_{k=0}^{\infty} \frac{1}{k!} B_{i,j}^k
$$
So in matrix form, using $\circ$ to denote the poitwise product of matrices,
$$
A = e^{-\lambda/2} \left( I + B + \frac{1}{2!} B\circ B + \frac{1}{3!} B\circ B\circ B + \cdots \right)
$$
Finally, we claim that this matrix is PSD.
First, note that the pointwise product of two PSD matrices is again PSD. Therefore, each of the terms $B\circ B\circ \cdots \circ B$ is PSD. The sum of PSD matrices is also PSD.
This is an infinite series of pointwise additions and products of PSD matrices. If this converges (which you know it does), it will also be PSD.
The answer is yes. With
$\text{rank}\big(A) = d$
What Sylvester's Law of Intertia tells you
is that your matrix is congruent to the $\text{n x n}$ matrix
$\begin{bmatrix} I_d & \mathbf 0\\ \mathbf 0 & \mathbf 0_{n-d}\mathbf 0_{n-d}^T\\ \end{bmatrix}$
More generally, any $\text{n x n}$ real symmetric matrix with rank of $d$ and $m$ positive eigenvalues is congruent to the $\text{n x n}$ matrix
$B =\begin{bmatrix} I_{m} & \mathbf 0&\mathbf 0\\ \mathbf 0 & -I_{d-m}&\mathbf 0\\ \mathbf 0 & \mathbf 0 & \mathbf 0\\ \end{bmatrix}$
where $(m,d-m)$ (or sometimes $(m,d-m, n-d)$) is the signature of a real symmetric matrix. Equivalently any $\text{n x n}$ real symmetric matrix with signature $(m,d-m)$ is an orbit of $B$.
Best Answer
It is true. The given conditions imply that $\mathcal X^2\preceq\|\mathcal X\|\mathcal X\preceq a\mathcal X$ in positive semidefinite partial ordering. Hence $E(\mathcal X^2)\preceq E(\|\mathcal X\|\mathcal X)\preceq E(a\mathcal X)$ and in turn, $\|E(\mathcal X^2)\|\le\big\|E\big(\|\mathcal X\|\mathcal X\big)\big\|\le\|E(a\mathcal X)\|$.