*Not yet an answer, work in progress*

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For conjecture 1, it is helpful to represent the determinant of an $n\times n$ matrix $M$ as an integral over anticommuting (<A HREF="https://en.wikipedia.org/wiki/Grassmann_number">Grassmann</A>) variables $\theta=(\theta_1,\theta_2,\ldots\theta_n)$, and their conjugates $\bar{\theta}=(\bar{\theta}_1,\bar{\theta}_2,\ldots\bar{\theta}_n)$,

$$\det M=\int d\theta\int d\bar{\theta}\,e^{\bar{\theta}\cdot M\cdot\theta}=\int d\theta\int d\bar{\theta}\,\prod_{i=1}^n\left(1+\bar{\theta}_i\sum_{j=1}^n M_{ij}\theta_j\right),\tag{1}$$
as explained, for example, in these <A HREF="http://ckw.phys.ncku.edu.tw/public/pub/Notes/PhaseTransitions/Zinn-Justin/QFT-RG/01._AlgebraicPreliminaries/1.7._GaussianIntegralsWithGrassmannVariables.pdf">lecture notes.</A>     

Apply this to $M=X^2+Y^2$,
$$\det(X^2+Y^2)=\int d\theta\int d\bar{\theta}\,e^{\bar{\theta}\cdot X^2\cdot\theta}e^{\bar{\theta}\cdot Y^2\cdot\theta}$$
$$\qquad=\int d\theta\int d\bar{\theta}\,\prod_{i,i'=1}^n\left(1+\bar{\theta}_i\sum_{j,k=1}^n X_{ik}X_{kj}\theta_j\right)\left(1+\bar{\theta}_{i'}\sum_{j',k'=1}^n Y_{i'k'}Y_{k'j'}\theta_{j'}\right).\tag{2}$$
We now take the expectation value over the independent normally distributed matrix elements of $X$ and $Y$,
$$\mathbb{E}[\det(X^2+Y^2)]=\int d\theta\int d\bar{\theta}\,\left(\mathbb{E}\biggl[\prod_{i=1}^n\biggl(1+\bar{\theta}_i\sum_{j,k=1}^n X_{ik}X_{kj}\theta_j\biggr)\biggr]\right)^2.\tag{3}$$

So I need to evaluate a Gaussian average of the form
$$Z_n=\mathbb{E}\biggl[\prod_{i=1}^n\biggl(1+\sum_{j,k=1}^n X_{ik}X_{kj}c_{ij}\biggr)\biggr]$$
and then perform the remaining integral of $Z_n^2$ over the coefficients $c_{ij}=\bar{\theta}_i\theta_j$,
with the help of the identities
$$\int d\theta_id\bar{\theta}_i=0,\;\;\int d\theta_id\bar{\theta}_i\,\bar\theta_i=0,\;\int d\theta_id\bar{\theta}_i\,\bar\theta_i\theta_i=1.$$

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As a quick check that this is leading somewhere, for $n=2$ one has
$Z_2=1+c_{11}+c_{22}+2c_{11}c_{22}$, giving
$$\mathbb{E}[X^2+Y^2]=\int d\theta\int d\bar{\theta}\,\biggl(1+\bar{\theta}_{1}\theta_1+\bar{\theta}_{2}\theta_2+2\bar{\theta}_{1}\theta_1\bar{\theta}_{2}\theta_2\biggr)^2$$
$$\qquad=\int d\theta\int d\bar{\theta}\,\biggl(1+2\bar{\theta}_{1}\theta_1+2\bar{\theta}_{2}\theta_2+6\bar{\theta}_{1}\theta_1\bar{\theta}_{2}\theta_2\biggr)=6,\tag{4}$$
which is the correct answer.
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