Two subgroup $A, B \le G$ are said to permute if $AB = BA$. A subgroup $H \le G$ is called $S$-semipermutable in $G$ if $H$ permutes with every Sylow $q$-subgroup of $G$ for primes $q$ not dividing $|H|$.
Theorem: Let $H$ be an $S$-semipermutable $\pi$-subgroup of $G$. Then $H^G$ contains a nilpotent $\pi$-complement, and all $\pi$-complements in $H^G$ are conjugate. Also, if $\pi$ consists of a single prime, then $H^G$ is solvable.
Corollary: If a Hall $\pi$-subgroup of $G$ is $S$-semipermutable, then $G$ contains a $\pi$-complement.
Proof: Write $M = H^G$. Then $M$ contains a $\pi$-complement $K$ by the above Theorem, and all $\pi$-complements of $M$ are conjugate. The Frattini argument thus applies, and we have $G = M N_G(K)$, so $|G : N_G(K)| = |M : N_G(K)|$ is a $\pi$-number, and so it suffices to find a $\pi$-complement in $N_G(K)$. This exists, however, since $N_G(K)$ is $\pi$-separable. To see this, observ that $K \subseteq N_M(K) \subseteq N_G(K)$ is a normal series in $N_G(K)$. Since $|K|$ is a $\pi$'-number, $|N_M(K) : K|$ is a $\pi$-number and $|N_G(K) : N_M(K)|= |G : M|$ is a $\pi'$-number, the result follows. $\square$
1) For the application of the Frattini argument, first $G$ acts on the $\pi$-complements $K$, and because by definition $H^G$ is generated by the conjugates of $H$, each conjugate of some $\pi$-complement $K$ is also in $M = H^G$. But why is the operation of $M$ on these conjugates transitive (which is necessary for the application of the Frattini argument), I guess because they are already conjugate in $M$, but this is not said in the Theoerm?
2) Why if $N_G(K)$ is $\pi$-separable, it implies that is has a $\pi$-complement in it. And why in the proof is $|N_M(K) : K|$ a $\pi$-number?
3) Where do we need that $M$ is a Hall $\pi$-subgroup?
I think you are assuming that all of the groups involved are finite, but it would be helpful if you said so, so that people do not waste time thinking about the case when $G$ is infinite.
I am not sure what your problem is with Question 1 - you seem to have answered it yourself. It is stated in both the theorem and in the proof of the corollary that all $\pi$-complements are conjugate in $M$.
For 2, you are presumably expected to know the result that a finite $\pi$-separable group has a $\pi$-complement. The proof is a straightforward induction, using the Schur-Zassenhaus Theorem. $|N_M(K):K|$ is a $\pi$-number because $K$ is a $\pi$-complement in $M$.
For 3, you use the fact that $H$ is a Hall $\pi$-subgroup to infer that $|G:M|$ is a $\pi$-number at the end of the proof.