Construct initial marking for marked graph so that the graph is live and safe
Why can't you find a live and safe marking?

Number the transitions from left to right $t_1,t_2,t_3,t_4,t_5$ and the places from left to right $p_1, \ldots, p_{12}$,

We need to find a firing sequence that is possible, fires all transitions at least once, has net effect zero, and never produces an unsafe marking.

It doesn't matter where we start: if the net is live, we can "cycle" any such firing sequence by moving its starting transition to the end or vice versa.

So we may as well start with the "thick" transition in the middle, $t_2$. For it to be enabled, its input places must be marked. For the firing sequence we'll construct to be safe, its output places must be empty. So let's fire it. Now, its input places are empty, and its output places are marked. We must now be able to fire $t_3,t_4,t_5$. This is possible iff exactly one of $p_9,p_{10},p_{12}$ was (and is) marked. We can then fire all three in a row. So let's pick $p_9$ as the one that was set. Then, after firing $t_2$, $t_3$ will be able to fire; firing it will enable $t_5$; firing that will enable $t_4$; firing that will put the token back into $p_9$.

After firing $t_3$, $p_4$ will be set, enabling $t_1$. The firing of $t_1$ and of $t_3,t_5,t_6$ are independent: they don't influence each other.

Having fired both $t_1$ and $t_3,t_5,t_6$, we're back at our initial marking. The net doesn't actually give us a choice: we must fire $t_3$ first, and then both $t_1$ and the sequence $t_5,t_6$ to enable $t_2$, and no alternative firing sequences exist.

So with the initial marking $p_2,p_6,p_8,p_9$, the only possible firing sequences of this net are given by the regular language $(t_2t_3(t_1|t_5t_4))^* = (t_2t_3(t_1t_5t_4 \cup t_5t_1t_4 \cup t_5t_4t_1))^*$. All of the reachable markings are safe.

By the way the example is drawn, there appears to be theory behind this example. This marked graph is special in that it is composed from three plain cycles (the ovals) by fusing some of their transitions. I'll bet Reisig's book proves that all marked graphs constructed in such a way are live and safe.

Your second question: what if we remove the place marked in yellow? Well, this only removes a constraint on the firing sequences: all existing firing sequences will continue to exist, but new ones may arise.

Indeed: after $t_4$  has fired, or in the initial marking above, $t_3$ can now already fire before $t_2$. However, this is merely a relaxing of the order in which transitions are fired, and it doesn't produce any unsafe or unlive markings.
So the net remains live and safe. I suppose you'll have to prove this.

I'll guess there is another lemma in the book proving that this is true in general: that removing places from a live and safe marked graph always leaves it live and safe. I haven't read the book, so I'm not sure.