First Locker: Presumably we do not try the same combination twice! The probability our first attempt fails is $\frac{9}{10}$. Given that the first attempt failed, the probability the second attempt fails is $\frac{8}{9}$. Given that the first two attempts failed, the probability the third attempt fails is $\frac{7}{8}$. And given $3$ failures in a row, the probability of a fourth is $\frac{6}{7}$. Multiply. Lots of cancellation, we get $\frac{6}{10}$.
The cancellation is trying to tell us that we missed a simpler argument. And we certainly did. Imagine listing the $10$ combinations at random in a row: we will try the first $4$ numbers in the list. Then we succeed precisely if the combination is among the first $4$ in the list. So the probability we succeed is $\frac{4}{10}$, and therefore the probability we fail is $\frac{6}{10}$. Or equivalently we fail if the combination is among the last $6$ in the list. The probability of this is $\frac{6}{10}$.
Second Locker: Let the mandatory waiting time between tries be $a$ units. If we succeed immediately, our waiting time was $0$. If we succeed on the second try, our total waiting time was $a$. If we succeed on the third try, our total waiting time was $2a$. Continue. Finally, if we succeed on the $10$-th try, our total waiting time was $9a$.
For any trial, the probability we succeed on that trial is $\frac{1}{10}$. This is because the right combination is equally likely to be at any place in our sequence of choices. Thus the expected total waiting time is $\frac{1}{10}(0)+\frac{1}{10}(a)+\cdots+\frac{1}{10}(9a).$
This simplifies to $\frac{a}{10}(1+2+\cdots+9)$, which is $\frac{9a}{2}$.