Canonically, I don't know of any 'short cuts' for problems such as this. In general it's just trial-and-error along with some deductive elimination and a lot of branch-and-bound search. I do recommend keeping them in the long-multiplication format, since it makes the bookkeeping a bit easier.
As an example, consider the last problem. Since N*B=N (mod 10), then either B=1 (which can't be the case because the result is longer than the multiplicand) or N has no multiplicative inverse mod 10. This immediately eliminates most of the possibilities for the pair (N, B), leaving you with either N=0 (in which case the same situation can be repeated with B and O), N=5 and B odd, or N even and B=6. Then the repetition of the digit O upon multiplication by B can be used to narrow down possibilities even further; etc.
Similar analyses can start to whittle down the other problems; in the first you have a similar situation for the last digits, along with the fact (by counting digits) that S*E must be less than 10, which narrows down the possibilties for both sharply. In the second, M can't be 1 or 2 for similar product-length reasons, and you can eliminate M=3 without too much effort: M=3 forces A=1 (by product-length reasons), and the only product 31X*31Y that can produce a last digit of 3 and have enough length is 317*319, whose product 101123 doesn't match the ASYLUM digit pattern.
Above and beyond this sort of thing, though, you're mostly stuck with trial-and-error; partial-product versions (where all of the long-multiplication terms 'below the line' that sum to the final product are shown) tend to be substantially easier because there's more opportunity for this sort of deduction.