Conditional expectation to de maximum $E(X_1\mid X_{(n)})$

Let $Y$ be the index of the maximum value: $X_Y=X_{(n)}$ (the event of ties is unimportant, it has zero probability). Let $Z=X_{(n)}$

Then

$$E(X_1\mid Z ,Y) = \begin{cases} Z & \text{ if } Y=1 \\ \frac{1}{2}Z & \text{ elsewhere} \\ \end{cases} $$

Then, applying the iterated expectations property, and because $P(Y=k)=1/n$:

$$E(X_1\mid Z)=E( E(X_1\mid Z , Y))=\frac{1}{n}Z+ \frac{n-1}{n}\frac{Z}{2} = \frac{n+1}{n}\frac{Z}{2} $$

Edit: this is not need to answer the question, but (from a comment) the full probability can be obtained (some abuse of notation follows) thus:

$$P(X_1\mid Z ) =\sum_{Y=1}^n P(X_1,Y\mid Z )\\ =\sum_{Y=1}^n P(X_1\mid Y,Z ) P(Y \mid Z)\\ =\delta(X_1 -Z)\frac{1}{n}+\sum_{Y=2}^n P(X_1\mid X_1<Z ) \frac{1}{n}\\ =\frac{1}{n}\delta(X_1 -Z)+ \frac{n-1}{n}U[0,Z] $$

So, yes $P(X_1\mid X_{(n)} )$ can be seen a mixing of a delta with a (truncated) uniform.

You're looking for $E(X_1\mid \max)$. The probability that $X_1=\max$ is $1/n$. So $$ E(X_1\mid \max) = \frac 1 n E(X_1\mid X_1=\max, \max) + \frac{n-1} n E(X_1\mid X_1\ne \max, \max). $$ The first term is easy to find: $E(X_1\mid X_1=\max,\max) = \max$, so we have $$ E(X_1\mid \max) = \frac 1 n \max + \frac{n-1} n E(X_1\mid X_1\ne \max, \max). $$ Then you need to show that the conditional distribution of $X_1$ given $\max$ and given the event that $X_1\ne\max$ is uniform on the interval from $0$ to $\max$.