The number of comparisons to find minimum and maximum simultaneously will be $3n/2-2$ if we put the value for n this will be 232
Now for the second one at first we have to find the largest element using the tournament search and number of comparison for this will be n-1.And each element is involved in comparison at most “logn” times.Second largest element will be one of the elements with which largest element is compared and largest element wins.So there will be at most “logn” number of elements.So the number of comparisons will be (n-1)+(logn-1) which n+logn-2.