and
$\mu_{Y|x} = \mu_{2} + \rho \cdot \frac{\sigma_2}{\sigma_1}\cdot (x-\mu_{1})$
and the variance
$\sigma^{2}_{Y|x} = \sigma^{2}_{2}(1-\rho^2)$
and the conditonal density of $X$ given $Y=y$ is a normal distribution with the mean
$\mu_{X|y} = \mu_{1} + \rho\cdot \frac{\sigma_{1}}{\sigma_{2}}\cdot (y-\mu_{2})$
and the variance
$\sigma^{2}_{X|y} = \sigma^{2}_{1}(1-\rho^2)$
Moreover, one must compare coefficients to each other for the bivariate normal density and the defintion above. Thus
$$\color{red}{(1)} \quad \frac{-1}{2(1-p^2)} =\frac{-1}{102}$$
$$\color{red}{(2)} \quad \left( \frac{x-\mu_1}{\sigma_{1})}\right)^2 =\left( \frac{x+2}{1} \right)^2$$
$$\color{red}{(3)} \quad \left( \frac{y-\mu_1}{\sigma_{1})}\right)^2 =\left( 4(y-1) \right)^2$$
$$\color{red}{(4)} \quad \left( \frac{2\rho \cdot (x-\mu_{1})}{(\sigma_{1})} \frac{ (y-\mu_{2})}{(\sigma_{2})}\right) =\left( 2.8(x+2)(y-1) \right)$$
Although from the information I have provided I cannot derive the solution for $\mu_{1}$ and $\mu_{2}$?