In the Cambridge HL Maths Statistics & Probability section there is a chapter on t-distributions, however as per usual they do not explain where to begin with most questions.
Assuming $T \sim t_7$ ($n=7$), solve $$P(-t < T) + P(0 < T) + P(t < T) + P(2t < T) = 1.75.$$
I began by breaking it into $$P(-t < T < 0) + 2P(0 < T <t) + 3P(t < T < 2t) + 4P(2t < T) = 1.75$$ following the areas that overlap. Now since $P(-t < T < 0) = P(0 < T <t)$, it becomes $$ 3P(0 < T <t) + 3P(t < T < 2t) + 4P(2t < T) = 1.75 \\ 3P(0 < T < 2t) + 4P(2t < T) = 1.75 \\ 3P(0 < T) + P(2t < T) = 1.75 $$ and since the distribution is symmetric about 0, $P(0 < T) = 0.5$ thus $$ P(2t < T) = 0.25 $$ giving $2t = t_7^{-1}(0.25)$ so $t = 0.359$ but in the book it is giving me a different answer. I haven't been able to find my mistake so I was hoping someone could help me. Thank you.
Your approach is correct till $P(2t<T)$. I have done some more transformations. $P(2t<T)=P(T<-2t)=0.25$. If I look at a table I found that $t_7^{-1}(0.25)=-t_7^{-1}(0.75)=-0.711142$. Thus $-2t=-0.711142\Rightarrow t\approx 0.3556$
The difference is about $0.34$ percentage points. It depends on the context if this gap makes any difference.