Let $\int_{-2}^{2.5} f(x) d x=7, \int_{-2}^{-0.5} f(x) d x=5, \int_{1}^{2.5} f(x) d x=7$. Find $\int_{-0.5}^{1} f(x) d x=$ and $\int_{1}^{-0.5}(7 f(x)-5) d x=\square$

Step 1 ：We know that \(\int_{-2}^{2.5} f(x) d x=7\), \(\int_{-2}^{-0.5} f(x) d x=5\), and \(\int_{1}^{2.5} f(x) d x=7\).

Step 2 ：We can use the property of integrals that the integral from a to b plus the integral from b to c equals the integral from a to c. So, \(\int_{-2}^{2.5} f(x) d x = \int_{-2}^{-0.5} f(x) d x + \int_{-0.5}^{1} f(x) d x + \int_{1}^{2.5} f(x) d x\).

Step 3 ：Substituting the given values, we get: \(7 = 5 + \int_{-0.5}^{1} f(x) d x + 7\).

Step 4 ：Solving for \(\int_{-0.5}^{1} f(x) d x\), we get \(\int_{-0.5}^{1} f(x) d x = 7 - 5 - 7 = -5\).

Step 5 ：Next, let's find \(\int_{1}^{-0.5}(7 f(x)-5) d x\).

Step 6 ：We can use the property of integrals that the integral from a to b of a function times a constant equals the constant times the integral from a to b of the function. So, \(\int_{1}^{-0.5}(7 f(x)-5) d x = 7\int_{1}^{-0.5} f(x) d x - 5\int_{1}^{-0.5} d x\).

Step 7 ：We know that \(\int_{1}^{-0.5} f(x) d x = -\int_{-0.5}^{1} f(x) d x = 5\) (since the integral from a to b equals the negative of the integral from b to a).

Step 8 ：And \(\int_{1}^{-0.5} d x = -0.5 - 1 = -1.5\).

Step 9 ：So, \(\int_{1}^{-0.5}(7 f(x)-5) d x = 7*5 - 5*(-1.5) = 35 + 7.5 = 42.5\).

Step 10 ：So, \(\boxed{\int_{-0.5}^{1} f(x) d x = -5}\) and \(\boxed{\int_{1}^{-0.5}(7 f(x)-5) d x = 42.5}\).