Solve $x^2\equiv -3\pmod {\!91}$ by CRT lifting roots $\!\bmod 13\ \&\ 7$ Question 1) Solve $$x^2\equiv -3\pmod {13}$$
I see that $x^2+3=13n$. I don't really know what to do? Any hints?
The solution should be $$x\equiv \pm 6 \pmod {13}$$
Question 2) $\ $ [note $\bmod 7\!:\ x^2\equiv -3\equiv 4\iff x\equiv \pm 2.\,$ Here we lift to $\!\!\pmod{\!91}\ $ -Bill] 
Given $x\equiv \pm 6 \pmod {13}$ and $x\equiv \pm 2 \pmod {7}$ find solutions $\pmod {91}$. I see that $91=13 \times 7$, does it mean I have to use chinese remainder theorem on 4 equations? If,so $x=6\times 13\times 7 \times 7\times (13\times 7 \times 7)^{-1}...$

The numbers are small, and one could find the answers without the full CRT machinery. But we will go ahead and use a "general" procedure. 
The idea is that to solve $x\equiv a\pmod{7}$, $x\equiv b\pmod{13}$, you use 
$$x\equiv (C)(13)(a)+(D)(7)(b)\mod{91},$$
where $C$ is the inverse of $13$ modulo $7$ and $D$ is the inverse of $7$ modulo $13$. We can see I think easily that $C=-1$ and $D=2$ will work, so we get
$$x\equiv -13a+14b\pmod{91}.\tag{1}$$
Now let us for example take $a=-2$ and $b=6$, one of your $4$ possibilities. That gives $x\equiv 110\equiv 19\pmod{91}$. 
