This is not an answer, but I'll try to prove the claims in the hint, and hopefully offer some helpful perspective.
First, a single wolf does not catch the sheep, even without the lion.
Let's say the speed of the animals is 1. The strategy of the sheep is to wait at the center $O$ until the wolf gets very close, say at $W_0$. It then chooses some $O'$ on the extension of the segment $OW_0$, and moves in the direction perpendicular to the line $O'O$ for $t_1$ amount of time to point $S_1$. Notice that the wolf can't catch the sheep while it's moving from $O$ to $S_1$.
When the sheep is at $S_1$, it notices that the wolf's at position $W_1$, and moves in the direction perpendicular to the line $O'S_1$ for for $t_2$ amount of time to point $S_2$. Of the two possible perpendicular directions, he chooses the one such that $\angle W_1S_1S_2\geq \pi/2$. Notice that the wolf can't catch the sheep while it's moving from $S_1$ to $S_2$.
At $S_2$, the sheep repeat the process above to arrive at $S_3$, then repeat again and again, ad infinitum to $S_{\infty}$. Notice that on no segment $S_iS_{i+1}$ can the wolf catch the sheep!
If the sheep lets $t_i=\Delta t/i$, then the total distance traveled by the sheep in this fashion is
$$
\Delta t\left (1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots\right)=\infty
$$
But the length of $O'S_n$ is
$$
\sqrt{{|O'O|}^2+\Delta t^2 \left[1+\left(\frac{1}{2}\right)^2+\ldots+\left(\frac{1}{n}\right)^2\right]}
$$
which when $n$ goes to infinity is just
$$
\sqrt{{|O'O|}^2+\Delta t^2\cdot\frac{\pi^2}{6}}
$$
By taking say $OS_1=0.1$ and $\Delta t=0.1$, the sheep ensures that the chase is confined in a very small area, while it takes forever for the wolf to catch him.
Second, two wolves always catch the sheep, if there's no lion to offer help.
The wolves first move themselves to the projections of the sheep on x and y axis. From there $W_1$ keeps it's projection on the x axis the same as the sheep's, and uses any extra speed in moving towards the sheep in the y direction. $W_2$ acts similarly.
Let $v_x$ and $v_y$ be the speeds of the projections. Since the initial distance between $W_2$ and $S$ is at most the radius of the arena $R$, if the sheep is to avoid being caught, the total length of time for which $v_x\geq \sqrt{2}/2$ can't be greater than $\sqrt{2}R$. Similar reasoning for $W_1$ shows that the total length of time for which $v_y\geq \sqrt{2}/2$ can't be greater than $\sqrt{2}R$, either. But notice that the animals move at speed 1 means that $v_x^2+v_y^2=1$, which means if $v_x\geq \sqrt{2}/2$ then $v_y\leq \sqrt{2}/2$, and vice versa. This means after the wolves moved themselves to the projections, the sheep must be caught in no more than $2\sqrt{2}R$ amount of time.
Third, one lion safeguards the sheep from two wolves
The lion choose to stay on the segment connecting one of the wolves and the center of the arena, and uses any extra speed to move radially outward towards the wolf, thus keeping it far away from the center. The other wolf can be handled by the sheep as shown in the first case above.
