Given a binomial distribution with mean \(\mu = 2\) and variance \(\sigma^2 = 1\), we have:
\[
\mu = np = 2
\]
and
\[
\sigma^2 = np(1-p) = 1.
\]
From \(\mu = np = 2\), we derive \(p = \frac{2}{n}\).
Substituting this into the variance equation yields:
\[
n \cdot \frac{2}{n} \cdot \left( 1 - \frac{2}{n} \right) = 1.
\]
Simplifying this equation leads to:
\[
2 \left( 1 - \frac{2}{n} \right) = 1 \quad \Rightarrow \quad 2 - \frac{4}{n} = 1 \quad \Rightarrow \quad \frac{4}{n} = 1 \quad \Rightarrow \quad n = 4.
\]
Substituting \(n = 4\) back into \(p = \frac{2}{n}\) gives:
\[
p = \frac{2}{4} = \frac{1}{2}.
\]
We now need to compute \(P(X>1)\). This can be expressed as:
\[
P(X>1) = 1 - P(X \leq 1).
\]
For a binomial distribution \(X \sim B(4, \frac{1}{2})\), \(P(X \leq 1)\) is calculated as:
\[
P(X \leq 1) = P(X = 0) + P(X = 1).
\]
Using the binomial probability mass function \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we find:
\[
P(X = 0) = \binom{4}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^4 = \frac{1}{16}
\]
and
\[
P(X = 1) = \binom{4}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^3 = \frac{4}{16}.
\]
Therefore,
\[
P(X \leq 1) = \frac{1}{16} + \frac{4}{16} = \frac{5}{16}.
\]
Finally, the probability \(P(X>1)\) is:
\[
P(X>1) = 1 - \frac{5}{16} = \frac{11}{16}.
\]
The probability that \(X\) is greater than 1 is \(\frac{11}{16}\).