if $a$ is a quadratic residue then $x^2\equiv a \mod p$ and $\left(\frac ap\right)=1$
as $a\equiv b \mod p$
$x^2\equiv b \mod p$ so $b$ is a quadratic residue as well so $\left(\frac bp\right)=1$ so $\left(\frac ap\right)=\left(\frac bp\right)$
if $a$ is not a quadratic residue then $\left(\frac ap\right)=-1$ as $a\equiv b \mod p$ so $b$ is also not a quadratic residue $\left(\frac bp\right)=-1$
so once again $\left(\frac ap\right)=\left(\frac bp\right)$
as for the second part whatever $a$ is if $x^2\equiv a^2 \mod p$ then $a^2$ is a quadratic residue so $\left(\frac {a^2}p\right)=1$
If $a\equiv b \mod p$
if $a$ is a quadratic residue then $x^2\equiv a \mod p$ and $\left(\frac ap\right)=1$
as $a\equiv b \mod p$
$x^2\equiv b \mod p$ so $b$ is a quadratic residue as well so $\left(\frac bp\right)=1$ so $\left(\frac ap\right)=\left(\frac bp\right)$
if $a$ is not a quadratic residue then $\left(\frac ap\right)=-1$ as $a\equiv b \mod p$ so $b$ is also not a quadratic residue $\left(\frac bp\right)=-1$
so once again $\left(\frac ap\right)=\left(\frac bp\right)$
as for the second part whatever $a$ is if $x^2\equiv a^2 \mod p$ then $a^2$ is a quadratic residue so $\left(\frac {a^2}p\right)=1$