Both are CFL
1) L = anbk ( n<=k<=2n)
for this language we can write the grammar as follows
S -> aSb | aSbb | $\epsilon$
2) L ={aibjck | (i≤j) or (j≤i), j=k}
i<=j or j<=i simply means i and j can be anything
So only one condition remains i.e. j=k ( number of b and c equal )
Hence it is CFL