1.
设函数\(f\left( x \right)=\left| x-a \right|\).
\((1)\)当\(a=2\)时,解不等式\(f\left( x \right)\geqslant 4-\left| x-1 \right|\);
\((2)\)若\(f\left( x \right)\leqslant 1\)的解集为\(\left[ 0,2 \right]\),\(\dfrac{1}{m}+\dfrac{1}{2n}=a\left( m > 0,n > 0 \right)\),求证:\(m+2n\geqslant 4\).