Apply Banach's fixed point theorem
A related problem.
You are doing fine, the $\sin(x)$ should be cancelled and you are left with
$$ ||Tf-Tg|| \leq \frac{2}{5}\int_0^1 |x^2+t^5||f(t)-g(t)| dt \leq \frac{4}{5}||f-g||_{\infty}\,$$
which proves your operator is an attractive operator on the Banach space you are given.
very easy ! $$|\frac 2 5\int_{0}^{1}(x^2+t^2)f(t)dt-\frac 2 5\int_{0}^{1}(x^2+t^2)g(t)dt|\leq|\frac 2 5\int_{0}^{1}(x^2+t^2)|f(t)-g(t)|dt| \\\leq \max|f(t)-g(t)|\frac 2 5\int_{0}^{1}(x^2+t^2)|dt|\leq\max|f(t)-g(t)|\frac 8 {15}$$