(a) $x≡5\pmod 7\;\;,\; x≡7\pmod{11}\;\;,\;\;x≡3\pmod{13}$

(b) $x≡3\pmod{10}\;\;,\;\; x≡8\pmod{15}\;\;,\;\;x≡5\pmod{84}$

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for (a) I have a rough idea how to do it, its like:

$n\_1=7,n\_2=11,n\_3=13$ then $n=7·11·13=1001$

${1001\over7}·k\_1≡1mod(7)$

${1001\over11}·k\_2≡1mod(11)$

${1001\over13}·k\_3≡1mod(13)$

Hence, $k\_1=[5], k\_2=[4], k\_3=[12]$

$x\_0=(11·13)·5·5+(7·13)·4·7+(7·11)·(-1)·3=287$

the solution set is $x=x\_0+k\_n, k\leftarrow\mathbb{Z}, x=887+k(1001)$

Im not sure if im correct, but i just follow the standard procedure

(b) for this one, what should I do the the $n$'s are not coprime, or prime???

Thank you!!